Let's Code_: LeetCode

Showing posts with label LeetCode. Show all posts

Thursday, September 1, 2022

interviewBit Medium: Palindrome Partitioning II

Problem Name: Palindrome Partitioning II

Problem Description: https://www.interviewbit.com/problems/palindrome-partitioning-ii/

Problem Approach used: This problem can be solved with MCM approach. You can note this when you feel the urge to partition the input String into parts (which are palindromes themselves in this case).

Time Complexity: O(n^2) worst-case time complexity and O(n^2) auxiliary space for memoisation.

Solution:

public class Solution {
    public int minCut(String A) {
        
        int memo[][] = new int[502][502];
        for (int i=0; i<502; i++) {
            Arrays.fill (memo[i], -1);
        }
        // i=0, j=n-1 // to ensure 2 partitions
        return palinPartition(A, 0, A.length()-1, memo);
    }
    
    int palinPartition(String A, int i, int j, int [][] memo) {
        if (i >= j) {
            return 0;
        }
        
        if (isPalindrome(A, i, j)) {
            return 0;
        }
        
        if (memo[i][j] != -1) {
            return memo[i][j];
        }
        
        int mn = Integer.MAX_VALUE;
        // k= i -> j-1
        for (int k=i; k<j; k++) {
            
            int c1 = palinPartition(A, i, k, memo);
            int c2 = palinPartition(A, k+1, j, memo);
            int temp = c1+c2+1;
            
            mn = Math.min (mn, temp);
        }
        
        return memo[i][j] = mn;
        
    }
    
    boolean isPalindrome(String s, int i, int j) {
        
        if (i >= j)
            return true;
            
        while (i<j) {
            if (s.charAt(i) != s.charAt(j)) {
                return false;
            } else {
                i++;
                j--;
            }
        }
        
        return true;
    }
}

Until Next time, Keep coding and have fun!

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Tuesday, July 19, 2022

Leetcode Medium: Container with most water

Problem Name: Container with most water

Problem Description: https://leetcode.com/problems/container-with-most-water

Problem Approach used: This problem can be solved with multiple approaches. The naive solution calculates the water content in all possible containers, compares and returns the maximum water that can be contained. It is O(n^2) time solution with O(1) auxiliary space.

Other "smarter" solution takes O(n) time with O(1) space. It makes use of the 2-pointer technique.

Time Complexity: O(n) worst-case time complexity and O(1) auxiliary space complexity.

Solution 1:

class Solution {
 public int maxArea(int[] height) {
     int maxArea = Integer.MIN_VALUE;
     for (int i=0; i<height.length; i++) {
         for (int j=i+1; j<height.length; j++) {
             int lesserHeight = Math.min(height[i], height[j]);
             maxArea = Math.max (maxArea, lesserHeight*(j-i));
         }
     }
     return maxArea;
 }
}

Solution 2:

class Solution {
    public int maxArea(int[] height) {
        // 2 pointer technique
        int smallerIndex=0, greaterIndex=height.length-1;
        int maxAreaSoFar = 0;
        while (smallerIndex < greaterIndex) {
            if (height[smallerIndex] >= height[greaterIndex]) {
                maxAreaSoFar = Math.max(maxAreaSoFar, 
                                        height[greaterIndex] * (greaterIndex - smallerIndex));
                greaterIndex-- ;
            } else if (height[smallerIndex] < height[greaterIndex]) {
                maxAreaSoFar = Math.max(maxAreaSoFar,
                                        height[smallerIndex] * (greaterIndex - smallerIndex));
                smallerIndex++;
            } else {
                ;
            }
        }
        return maxAreaSoFar;
    }
}

Until Next time, Keep coding and have fun!

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Monday, July 18, 2022

Leetcode Hard: Trapping Rain Water

Problem Name: Trapping Rain Water

Problem Description: https://leetcode.com/problems/trapping-rain-water/

Problem Approach used: Calculate the amount of water this index would store with the knowledge of highest left and highest right tower. Then subtract the hight of the current tower as that's just conctrete, water can't seep into.

Time Complexity: O(n) worst-case time and space complexity.

Solution:

class Solution {
    public int trap(int[] height) {
        int n = height.length;
        int[] leftMax = new int[n];
        int[] rightMax = new int[n];
        
        int maxHt = 0;
        for (int i=0; i<n; i++) {
            maxHt = Math.max (maxHt, height[i]);
            leftMax[i] = maxHt;
        }
        
        maxHt = 0;
        for (int i=n-1; i>=0; i--) {
            maxHt = Math.max (maxHt, height[i]);
            rightMax[i] = maxHt;
        }
        
        int totalWater = 0;
        int water = 0;
        for (int i=0; i<n; i++) {
            water = Math.min(leftMax[i], rightMax[i]) - height[i];
            totalWater += water;
        }
        
        return totalWater;
    }
}

Keep coding and have fun!

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Saturday, May 7, 2022

LeetCode Medium: Product of Array Except Self

Problem Name: Product of Array Except Self

Problem Description: https://leetcode.com/problems/product-of-array-except-self/

Problem Approach used: It's an easy problem to solve, just keep in mind your Indeterminate Numbers lesson of High School to solve it.

Time Complexity: O(n) worst-case time complexity and O(1) auxiliary space complexity.

Solution:

class Solution {
    public int[] productExceptSelf(int[] nums) {
        long prodArr = 1;
        long non0ProdArr = 1;
        int countZeroes = 0;
        for (int i=0; i<nums.length; i++) {
            if (nums[i] == 0)
                countZeroes++;
            if (nums[i] != 0)
                non0ProdArr *= nums[i];
            prodArr *= nums[i];
        }
        int[] answer = new int[nums.length];
        if (countZeroes > 1) {
            Arrays.fill(answer, 0);
            return answer;
        }
        
        for (int i=0; i<nums.length; i++) {
            if (nums[i] == 0 && countZeroes == 1) {
                    countZeroes--;
                    answer[i] = (int)non0ProdArr;
            } else {
                answer[i] = (int)(prodArr / (long)nums[i]);
            }
        }
        
        return answer;
    }
}

Until Next time, Keep coding and have fun!

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Tuesday, April 5, 2022

Leetcode Medium: Set Matrix Zeroes

Problem Name: Set Matrix Zeroes

Problem Description: https://leetcode.com/problems/set-matrix-zeroes/

Problem Approach used: Its a trick problem to solve it in constant space. We've used the same, using HashSets in the below solution.

Time Complexity: O(m*n) worst-case time complexity and O(1) auxiliary space complexity.

Solution:

// In 1 line of thought, constant space solution is ideally not possible as all possible int values can be part of the matrix, but if we can assume some sentinel value out of these (like Integer.MIN_VALUE in our case) for marking original zeroes, a solution follows.

// * Also we can use a trick to solve ths problem: marking 1st(top and left) elements of the row and column respetively, to 0.

class Solution {
    Set<Integer> toZeroRows = new HashSet<>(), toZeroColumns = new HashSet<>();
    
    public void setZeroes(int[][] matrix) {
        for (int row=0; row<matrix.length; row++) {
            for (int column=0; column<matrix[0].length; column++) {
                if (matrix[row][column] == 0) {
                    toZeroRows.add(row);
                    toZeroColumns.add(column);
                }
            }
        }
        // Print initial matrix
        printMatrix(matrix);
        
        for (int row=0; row<matrix.length; row++) {
            if (toZeroRows.contains(row)) {
                for(int columnIndex=0; columnIndex<matrix[0].length; columnIndex++) {
                    matrix[row][columnIndex] = 0;
                }
            }
        }
        
        printMatrix(matrix);
        
        for (int column=0; column<matrix[0].length; column++) {
            if (toZeroColumns.contains(column))
                for(int rowIndex=0; rowIndex<matrix.length; rowIndex++) {
                    matrix[rowIndex][column] = 0;
                }
        }
        printMatrix(matrix);
        
    }
    
    public void printMatrix(int [][] matrix) {
        for (int row = 0; row<matrix.length; row++) {
            for (int column = 0; column<matrix[0].length; column++) {
                System.out.print(" " + matrix[row][column]);
            }
            System.out.println();
        }
    }
}

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Leetcode Problem: Climbing Stairs

Problem of Today: Climbing Stairs

Problem description: https://leetcode.com/problems/climbing-stairs/

Solution Approach: Memoization

Solution:

class Solution {
    private int memo[];
    
    // Bottom-up
    public int climbStairs(int n) {
        memo = new int[n+1];
        for (int i=0; i<=n; i++) {
            memo[n] = -1;
        }
        memo[0] = 1;
        memo[1] = 1;
        for (int i=2; i<=n; i++) {
            memo[i] = memo[i-1] + memo[i-2];
        }
        
        return memo[n];
    }
}

Happy Coding!

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Monday, December 6, 2021

LeetCode Medium: Generate Parentheses

Dear followers,

Tonight, I came across a nice problem for post-dinner exercise, about generating all possible valid parentheses sequences given the number of brackets to use in total.

PROBLEM LINK: https://leetcode.com/problems/generate-parentheses/

PROBLEM SOLVING APPROACH: Recursion (using Choice Diagram and Decision tree)

TIME-COMPLEXITY: O(2^n) in the worst case but the input n is given to be max 8, so very much doable :)

Here's a sample Java solution for your reference:

class Solution {
    public List<String> generateParenthesis(int n) {
        ArrayList<String> generatedParentheses = new ArrayList<>();
        genParenthesesRecur(n, n, "", generatedParentheses);
        return generatedParentheses;
    }
    
    private void genParenthesesRecur (int remOpenBrackets, int remClosingBrackets, String outputFromCaller, List<String> generatedParentheses) {
        // Base Cases
        if (remOpenBrackets < 0 || remClosingBrackets < 0)
            return;
        
        if (remClosingBrackets < remOpenBrackets) {
            return;
        }
        
        if (remOpenBrackets == 0 && remClosingBrackets == 0) {
           generatedParentheses.add(outputFromCaller);
        }
        
        // Recursive Case
        String opUsingAnOpeningBracket = outputFromCaller + "(";
        genParenthesesRecur(remOpenBrackets-1, remClosingBrackets, opUsingAnOpeningBracket, generatedParentheses);
        if (remClosingBrackets > remOpenBrackets) {
            String opUsingAClosingBracket = outputFromCaller + ")";
            genParenthesesRecur(remOpenBrackets, remClosingBrackets-1, opUsingAClosingBracket, generatedParentheses);
        }
    }
}

Thanks for Reading!

Happy Programming!!

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Sunday, December 5, 2021

Leetcode Easy: Longest Common Prefix

Tea-time snacking!

PROBLEM: https://leetcode.com/problems/longest-common-prefix/

TIME-COMPLEXITY: O(n*m) where n is the number of strings in input and m is the length of longest common prefix.

Sample Java Solution for tea-time snacking:

class Solution {
    public String longestCommonPrefix(String[] strs) {
        if (strs.length == 1)
            return strs[0];
        
        int minLen = Integer.MAX_VALUE;
        for (int i=0; i<strs.length; i++) {
            minLen = Math.min(minLen, strs[i].length());
        }
        int lastPastIdx = minLen;
        outer:
        for (int i=0; i<minLen; i++) {
            char ch = strs[0].charAt(i);
            for (int str=0; str<strs.length; str++) {
                if (strs[str].charAt(i) != ch) {
                    lastPastIdx = i;
                    break outer;
                }
            }
        }
        return strs[0].substring(0, lastPastIdx);
    }
}

Enjoy and Happy Coding!!

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Saturday, December 4, 2021

LeetCode Medium Trick Problem: Longest Consecutive Sequence

Came across this problem on Leetcode, which requires one to focus on optimization.

PROBLEM LINK: https://leetcode.com/problems/longest-consecutive-sequence

TIME COMPLEXITY: O(n) at worst case where n is the number of elements in the input array (precisely the constant factor for n is 2)

APPROACH: The most naive approach to solve this problem would have been to sort the input array using comparison sort in O(n lg n) worst case time complexity and then in single pass find the length of the consecutive sequences in the same, resulting in overall worst case time complexity on O(n lg n).

But, if we notice carefully, we can make use of a trick to solve the problem in O(n) worst case time complexity. The trick makes a space trade off of O(n) and makes use of a HashSet to store the values to quickly check if any number exists in the input array. Then, it checks for all values in input, if it is the start of a consecutive sequence by checking presence of 1 lesser value than that on the number line, in the number Set. If any start-of-consecutive-range is found in the array elements, the length of the range is found by consecutively checking all the numbers in the increasing order on the number line consecutively.

In each such iteration of the numbers in input array, the length of the max-range is updated and the largest value is returned at the end of all iterations.

Here's a sample code for the same:

class Solution {
    
    public int longestConsecutive(int[] nums) {
        Set numberSet = new HashSet ();
        
        // Add numbers to Set while updating length of the longest consecutive elements sequence
        int maxLLCES = 0, currLLCES = 0;
        for (int i=0; i<nums.length; i++) {
            numberSet.add (nums[i]);
        }
        
        for (int i=0; i<nums.length; i++) {
            if (!numberSet.contains(nums[i]-1)) {
                // nums[i] is not a start of a consecutive sequence
                currLLCES = 1;
                int checkNum = nums[i]+1;
                while(numberSet.contains(checkNum)) {
                    currLLCES++;
                    checkNum++;
                }
                maxLLCES = Math.max(maxLLCES, currLLCES);
                // System.out.println("Updated maxLLCES:" + maxLLCES);
            }
        }
        return maxLLCES;
    }
}

Do share your thoughts and feel free to talk about the alternatives/optimisations you feel can be done in the comment section!

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Friday, December 3, 2021

Recursion Series: Deleting the middle element from a stack

This post marks the start point of the much awaited recursion series on Let'sCode_ =)

We start it with a GeeksForGeeks problem : https://www.geeksforgeeks.org/delete-middle-element-stack/

Position of Middle element of all elements from top of stack (1 based):

stack.size()/2 + 1

Now, to build an effective solution we can use:

Methodology: Recursion

Time Complexity: O(n) where n is the number of elements in the stack

Space Complexity: O(n) considering the accumulation of constant space in each level of the recursive stack. O(1) if we dis-consider the stack space accumulation.

Here is one such solution:

<-- --todeletepos="" base="" case="" deletemidrecur="" int="" nteger="" printstack="" private="" return="" stack.pop="" stack.push="" stack="" static="" tack="" top="" void="">/*package whatever //do not write package name here */

import java.io.*;
import java.util.*;

class GFG {
	public static void main (String[] args) {
		System.out.println("GfG!");
		Stack<Integer> stack = new Stack<>();
		
		stack.push(6);
		stack.push(5);
		stack.push(4);
		stack.push(3);
		stack.push(2);
		stack.push(1);
		deleteMid(stack);
		
	}
	
	private static void deleteMid(Stack<Integer> stack) {
	    if (stack == null)
	        return;
	    if (stack.isEmpty())
	        return;
	        
	    int mid = stack.size()/2+1;
	    deleteMidRecur(stack, mid);
	    printStack(stack);
	}
	
	private static void deleteMidRecur (Stack<Integer> stack, int toDeletePos) {
	    if (toDeletePos == 1) {
	        // delete this element <-- base case
	        stack.pop();
	        return;
	    }
	    
	    int top = stack.pop();
	    deleteMidRecur(stack, --toDeletePos);
	    stack.push(top);
	}
	
	private static void printStack (Stack<Integer> stack) {
	    while (!stack.isEmpty()) {
	        System.out.println(stack.pop());
	    }
	}
}

IMO, such are the most efficient solutions to this problem. So share your thoughts and let me know your point of views.

Happy Coding!

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Sunday, August 29, 2021

LeetCode Medium: Course Schedule

The next problem in the current coding spree was:

Problem Name: Course Schedule

Problem Description: https://leetcode.com/problems/course-schedule/

Problem Approach used: Detecting cycles on the directed graph pf dependencies using DFS to solve in O(V+E) where V is the number of courses in the input and E is the number of edges between them.

Time Complexity: O(lg n) worst-case time and O(1) auxiliary space complexity

Java Solution:

//Cycle detection

// package com.projects.cv.course_schedule;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Solution {

    // private static Logger logger = Logger.getLogger("MyLogger");
    int nodes;


    public boolean canFinish(int numCourses, int[][] prerequisites) {

        if (prerequisites == null)
            return false;
        int pL = prerequisites.length;
        nodes = numCourses;

        // Build directed graph
        Map<Integer, List<Integer>> g = new HashMap<>();
        for (int i=0; i<pL; i++) {
            if (!g.containsKey(prerequisites[i][0]))
                g.put(prerequisites[i][0], new ArrayList<>());
            g.get(prerequisites[i][0]).add(prerequisites[i][1]);
        }

        System.out.println(g);

        //Create visited to prevent re-visiting
        int visited[] = new int[numCourses];

        for (int i=0; i<numCourses; i++) {
            if (visited[i] == 0) {
                if (hasCycleDfs(g, visited, i, -1))
                    return false;
            }
        }

        return true;

    }

    private boolean hasCycleDfs( Map<Integer, List<Integer>> g, int[] visited, int n, int parent) {
        if (visited[n] == -1) {
            //current exploration path
            System.out.println("cycle found at u(" + parent + ")->v(" + n + ")");
            return true;
        }
        if (visited[n] == 1) {
            return false;
        }

        visited[n] = -1;

        if (g.get(n) == null) { // tackle bad callers
            visited[n] = 1;
            return false;
        }

        for (int neighBr : g.get(n)) {
            if (hasCycleDfs(g, visited, neighBr, n))
                return true;
        }

        visited[n] = 1;

        return false;
    }

    // public static void main(String[] args) {
    //     Solution s = new Solution();
    //     int[][] deps = {{1, 0}};
    //     s.canFinish(2, deps);
    // }

}

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GeeksforGeeks Medium: Find the Number of Islands

Yesterday I solved a few hands-on coding problems using Java programming language.

I came across this easy problem(called Medium there) over GfG, to turn on the inertia:

Problem: Find the Number of Islands

Problem Description: https://practice.geeksforgeeks.org/problems/find-the-number-of-islands

Problem Approach: DFS

Time Complexity: O(V) where v = number of cells in the input grid

One Java Solution:

// { Driver Code Starts
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG
{
    public static void main(String[] args) throws IOException
    {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int T = Integer.parseInt(br.readLine().trim());
        while(T-->0)
        {
            String[] s = br.readLine().trim().split(" ");
            int n = Integer.parseInt(s[0]);
            int m = Integer.parseInt(s[1]);
            char[][] grid = new char[n][m];
            for(int i = 0; i < n; i++){
                String[] S = br.readLine().trim().split(" ");
                for(int j = 0; j < m; j++){
                    grid[i][j] = S[j].charAt(0);
                }
            }
            Solution obj = new Solution();
            int ans = obj.numIslands(grid);
            System.out.println(ans);
        }
    }
}// } Driver Code Ends



class Solution
{
    int r, c;
    byte[] xOffset = {1, 1, 1, 0, -1, -1, -1, 0};
    byte[] yOffset = {1, 0, -1, -1, -1, 0, 1, 1};
    
    //Function to find the number of islands.
    public int numIslands(char[][] grid)
    {
        // Code here
        r = grid.length;
        c = grid[0].length;
        
        boolean[][] visited = new boolean[r][c];
        int cnt = 0;
        for (int i=0; i<r; i++) {
            for(int j=0; j<c; j++) {
                if (grid[i][j]=='1' && !visited[i][j]) {
                    dfs(grid, visited, i, j);
                    cnt++;
                }
            }
        }
        
        return cnt;
    }
    
    private void dfs (char[][] grid, boolean[][] visited, int x, int y) {
        //input validation
        if (!valid (grid, x, y) || visited[x][y]==true) {
            return;
        }
        
        visited[x][y] = true;
        
        for (int i=0; i<8; i++) {
            int newX = x + xOffset[i];
            int newY = y + yOffset[i];
            
            if(valid(grid, newX, newY) && !visited[newX][newY]) {
                dfs(grid, visited, newX, newY);
            }
        }
    }
    
    boolean valid (char[][]grid, int x, int y) {
        if (x<0 || x>=r || y<0 || y>=c || grid[x][y] == '0'){
            return false;
        }
        return true;
    }
}

Do share your thoughts and feel free to talk about the alternatives/optimisations you feel can be done in the comment section!

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Saturday, January 23, 2021

LeetCode Medium: Open the Lock

Another BFS solvable problem of LeetCode!

Problem Description: https://leetcode.com/problems/open-the-lock/

Problem Approach: BFS

Time Complexity: O(1) as the custom size of state-nodes in this problem is a constant.

Solution:

//public class OpenTheLock {
//}

import java.util.*;

class State {
    String code;
    int dist;
    State(String code, int dist) {
        this.code = code;
        this.dist = dist;
    }
}

class Solution {
    public int openLock(String[] deadends, String target) {

        Set<String> visited = new HashSet<>();
        Set<String> deadendsSet = new HashSet<>();
        for (String d : deadends) {
            deadendsSet.add(d);
        }

        if ("0000".equals(target))
            return 0;
        if (deadendsSet.contains("0000"))
            return -1;

        Queue<State> q = new LinkedList<>();
        q.add(new State("0000", 0));
        visited.add("0000");
        while (!q.isEmpty()) {
            State out = q.remove();
            List<String> neighbours = getNeighbours(out.code);
            for (String n : neighbours) {
                if (!visited.contains(n) && !deadendsSet.contains(n)) {
                    if (n.equals(target))
                        return out.dist+1;
                    State neighState = new State(n, out.dist+1);
                    visited.add(n);
                    q.add(neighState);
                }
            }
        }
        return -1;
    }

    private List<String> getNeighbours(String baseState) {
        List<String> res = new ArrayList<>();
        char[] codeChar = baseState.toCharArray();
        for (int i=0; i<codeChar.length; i++) {
            String neighState = new String();

            String c = ((codeChar[i]-'0')+1)%10 + "";
            for (int idx=0; idx<codeChar.length; idx++)
                if (idx == i)
                    neighState+=c;
                else
                    neighState+=codeChar[idx];
            res.add(neighState);

            neighState = new String();
            c = ((codeChar[i]-'0')+10-1)%10 + "";
            for (int idx=0; idx<codeChar.length; idx++)
                if (idx == i)
                    neighState+=c;
                else
                    neighState+=codeChar[idx];
            res.add(neighState);
        }
        return res;
    }

}

Have fun, until next time!

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LeetCode Medium: Snakes and Ladders

Here's a problem based on the classical snakes and ladders children's game

Problem Description: https://leetcode.com/problems/snakes-and-ladders

Solution Approach: BFS

Time Complexity: O(V) where V is the number of cells in the board-game and E = 6V.

Solution:

class Solution {
    public int snakesAndLadders(int[][] board) {
        int rows = board.length;
        int cols = board[0].length;
        int maxSq = rows*cols;
        
        boolean visited[] = new boolean[maxSq];
        int[] pred = new int[maxSq+1]; // Allows to print the path as well. Alternatively, a dist[] incrementing at each time can be used too.
        Queue<Integer> squaresQ = new LinkedList<>();
        squaresQ.add(1);
        visited[1] = true;
        pred[1] = 0;
        
        outer:
        while(!squaresQ.isEmpty()) {
            int sqOut = squaresQ.remove();
            
            for (int i=1; i<=6; i++) {
                int neighbourSq = sqOut + i;
                if (neighbourSq > maxSq)
                    break;
                
                int effRow = getEffRow(neighbourSq, rows, cols);
                int effCol = getEffCol(neighbourSq, rows, cols);
                if (board[effRow][effCol] != -1) {
                    neighbourSq = board[effRow][effCol];
                    effRow = getEffRow(neighbourSq, rows, cols);
                    effCol = getEffCol(neighbourSq, rows, cols);
                }
                if (neighbourSq == maxSq) {
                    pred[neighbourSq] = sqOut;
                    break outer;
                }
                
                if (!visited[neighbourSq]) {
                    visited[neighbourSq] = true;
                    pred[neighbourSq] = sqOut; // Alternatively, can do: dist[neighbourSq] = dist[sqOut]++;
                    squaresQ.add(neighbourSq);
                }
                
            }
        }
        
        int cnt = 0;
        int predSq = maxSq;
        if (pred[predSq] == 0)
            return -1;
        while(predSq != 1) {
            cnt++;
            predSq = pred[predSq];
        }
        
        return cnt;
    }
    
    private int getEffRow(int sq, int rows, int cols) {
       return (rows - 1 - (sq-1)/cols);
    }
    
    private int getEffCol(int sq, int rows, int cols) {
        int effCol = (sq -1) % cols; // 0 based
        
        if (((sq - 1)/cols + 1) % 2 == 0) // even row from bottom:1 based
            effCol = cols - 1 - effCol;
        
        return effCol;
    }
}

Until next time,

Happy Coding!

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