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:

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// 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: 

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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]; } }

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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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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;
    }
}

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