Given a binary matrix mat[][] containing 0’s and 1’s. Each row of the matrix is sorted in the non-decreasing order, the task is to find the left-most column of the matrix with at least one 1 in it.
Note: If no such column exists return -1.
Examples:
Input:
mat[][] = {{0, 0, 0, 1}
{0, 1, 1, 1}
{0, 0, 1, 1}}
Output: 2
Explanation:
The 2nd column of the
matrix contains atleast a 1.
Input:
mat[][] = {{0, 0, 0}
{0, 1, 1}
{1, 1, 1}}
Output: 1
Explanation:
The 1st column of the
matrix contains atleast a 1.
Input:
mat[][] = {{0, 0}
{0, 0}}
Output: -1
Explanation:
There is no such column which
contains atleast one 1.
Approach:
- Here we start our traversal from the last element of the first row. This includes two steps.
- If the current iterating element is 1, we decrement the column index. As we find the leftmost column index with value 1, so we don’t have to check elements with greater column index.
- If the current iterating element is 0, we increment row index. As that element is 0, we don’t need to check previous elements of that row.
- We continue until one of the row or column index become invalid.
Below is the implementation of the above approach.
C++
// C++ program to calculate leftmost // column having at least a 1#include <bits/stdc++.h>using namespace std;#define N 3#define M 4// Function return leftmost // column having at least a 1int FindColumn(int mat[N][M]){ int row = 0, col = M - 1; int flag = 0; while (row < N && col >= 0) { // If current element is // 1 decrement column by 1 if (mat[row][col] == 1) { col--; flag = 1; } // If current element is // 0 increment row by 1 else { row++; } } col++; if (flag) return col + 1; else return -1;}// Driver codeint main(){ int mat[N][M] = { { 0, 0, 0, 1 }, { 0, 1, 1, 1 }, { 0, 0, 1, 1 } }; cout << FindColumn(mat); return 0;} |
Java
// Java program to calculate leftmost // column having at least a 1import java.util.*;class GFG{static final int N = 3;static final int M = 4;// Function return leftmost // column having at least a 1static int FindColumn(int mat[][]){ int row = 0, col = M - 1; int flag = 0; while(row < N && col >= 0) { // If current element is // 1 decrement column by 1 if(mat[row][col] == 1) { col--; flag = 1; } // If current element is // 0 increment row by 1 else { row++; } } col++; if (flag!=0) return col + 1; else return -1;}// Driver codepublic static void main(String[] args){ int[][] mat = { { 0, 0, 0, 1 }, { 0, 1, 1, 1 }, { 0, 0, 1, 1 } }; System.out.print(FindColumn(mat));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 program to calculate leftmost# column having at least a 1N = 3M = 4# Function return leftmost# column having at least a 1def findColumn(mat: list) -> int: row = 0 col = M - 1 while row < N and col >= 0: # If current element is # 1 decrement column by 1 if mat[row][col] == 1: col -= 1 flag = 1 # If current element is # 0 increment row by 1 else: row += 1 col += 1 if flag: return col + 1 else: return -1# Driver Codeif __name__ == "__main__": mat = [ [0, 0, 0, 1], [0, 1, 1, 1], [0, 0, 1, 1] ] print(findColumn(mat))# This code is contributed by sanjeev2552 |
C#
// C# program to calculate leftmost // column having at least 1using System;class GFG{static readonly int N = 3;static readonly int M = 4;// Function return leftmost // column having at least a 1static int FindColumn(int [,]mat){ int row = 0, col = M - 1; int flag = 0; while(row < N && col >= 0) { // If current element is // 1 decrement column by 1 if (mat[row, col] == 1) { col--; flag = 1; } // If current element is // 0 increment row by 1 else { row++; } } col++; if (flag != 0) return col + 1; else return -1;}// Driver codepublic static void Main(String[] args){ int[,] mat = { { 0, 0, 0, 1 }, { 0, 1, 1, 1 }, { 0, 0, 1, 1 } }; Console.Write(FindColumn(mat));}}// This code is contributed by Rohit_ranjan |
Javascript
<script> // JavaScript program to calculate leftmost // column having at least 1 let N = 3; let M = 4; // Function return leftmost // column having at least a 1 function FindColumn(mat) { let row = 0, col = M - 1; let flag = 0; while(row < N && col >= 0) { // If current element is // 1 decrement column by 1 if (mat[row][col] == 1) { col--; flag = 1; } // If current element is // 0 increment row by 1 else { row++; } } col++; if (flag != 0) return col + 1; else return -1; } let mat = [ [ 0, 0, 0, 1 ], [ 0, 1, 1, 1 ], [ 0, 0, 1, 1 ] ]; document.write(FindColumn(mat));</script> |
Output:
2
Time Complexity: O(N + M). where N is number of row and M is number of column.
Space Complexity: O(1)
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