Leftmost Column with atleast one 1 in a row-wise sorted binary matrix

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 exist return -1.

Examples:

Input: 
mat[2][2] = { {0, 0},
              {1, 1} }
Output: 1
Explanation: 
The 1st column of the
matrix contains atleast a 1.

Input: 
mat[2][2] = {{0, 0},
             {0, 0}}
Output: -1
Explanation:
There is no such column which 
contains atleast one 1.

Approach: The idea is to iterate over all the rows of the matrix and Binary Search over them for the first occurrence of the 1 in the that row. The minimum of all the occurrence of the first 1 in the rows of the matrix will be the desired solution for the problem.

Below is the implementation of the above approach:



C++

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// C++ implementation to find the
// Leftmost Column with atleast a 
// 1 in a sorted binary matrix
  
#include <bits/stdc++.h>
  
#define N 3
using namespace std;
  
// Function to search for the 
// leftmost column of the matrix
// with atleast a 1 in sorted 
// binary matrix
int search(int mat[][3], int n, int m)
{
    int i, a = INT_MAX;
      
    // Loop to iterate over all the 
    // rows of the matrix
    for (i = 0; i < n; i++) {
        int low = 0;
        int high = m - 1;
        int mid;
        int ans = INT_MAX;
          
        // Binary Search to find the 
        // leftmost occurence of the 1
        while (low <= high) {
            mid = (low + high) / 2;
              
            // Condition if the column 
            // contains the 1 at this 
            // position of matrix
            if (mat[i][mid] == 1) {
  
                if (mid == 0) {
                    ans = 0;
                    break;
                }
                else if (mat[i][mid - 1] == 0) {
                    ans = mid;
                    break;
                }
            }
              
            if (mat[i][mid] == 1)
                high = mid - 1;
            else
                low = mid + 1;
        }
          
        // If there is a better solution
        // then update the answer
        if (ans < a)
            a = ans;
    }
      
    // Condition if the solution 
    // doesn't exist in the matrix
    if (a == INT_MAX)
        return -1;
    return a+1;
}
  
// Driver Code
int main()
{
    int mat[3][3] = { 0, 0, 0,
                    0, 0, 1,
                    0, 1, 1 };
    cout << search(mat, 3, 3);
    return 0;
}

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Java

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// Java implementation to find the
// Leftmost Column with atleast a 
// 1 in a sorted binary matrix
import java.util.*;
  
class GFG{
      
static final int N = 3
  
// Function to search for the 
// leftmost column of the matrix
// with atleast a 1 in sorted 
// binary matrix
static int search(int mat[][], int n, int m)
{
    int i, a = Integer.MAX_VALUE;
      
    // Loop to iterate over all the 
    // rows of the matrix
    for(i = 0; i < n; i++)
    {
       int low = 0;
       int high = m - 1;
       int mid;
       int ans = Integer.MAX_VALUE;
         
       // Binary Search to find the 
       // leftmost occurence of the 1
       while (low <= high)
       {
           mid = (low + high) / 2;
             
           // Condition if the column 
           // contains the 1 at this 
           // position of matrix
           if (mat[i][mid] == 1)
           {
               if (mid == 0)
               {
                   ans = 0;
                   break;
               }
               else if (mat[i][mid - 1] == 0
               {
                   ans = mid;
                   break;
               }
           }
           if (mat[i][mid] == 1)
               high = mid - 1;
           else
               low = mid + 1;
       }
         
       // If there is a better solution
       // then update the answer
       if (ans < a)
           a = ans;
    }
      
    // Condition if the solution 
    // doesn't exist in the matrix
    if (a == Integer.MAX_VALUE)
        return -1;
    return a + 1;
}
  
// Driver Code
public static void main(String[] args)
{
    int mat[][] = { { 0, 0, 0 },
                    { 0, 0, 1 },
                    { 0, 1, 1 } };
                      
    System.out.print(search(mat, 3, 3));
}
}
  
// This code is contributed by Amit Katiyar

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C#

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// C# implementation to find the leftmost  
// column with atleast a 1 in a sorted
// binary matrix
using System;
  
class GFG{
      
//static readonly int N = 3; 
  
// Function to search for the leftmost  
// column of the matrix with atleast a
//  1 in sorted binary matrix
static int search(int [,]mat, int n, int m)
{
    int i, a = int.MaxValue;
      
    // Loop to iterate over all the 
    // rows of the matrix
    for(i = 0; i < n; i++)
    {
       int low = 0;
       int high = m - 1;
       int mid;
       int ans = int.MaxValue;
         
       // Binary Search to find the 
       // leftmost occurence of the 1
       while (low <= high)
       {
           mid = (low + high) / 2;
             
           // Condition if the column 
           // contains the 1 at this 
           // position of matrix
           if (mat[i, mid] == 1)
           {
               if (mid == 0)
               {
                   ans = 0;
                   break;
               }
               else if (mat[i, mid - 1] == 0) 
               {
                   ans = mid;
                   break;
               }
           }
           if (mat[i, mid] == 1)
               high = mid - 1;
           else
               low = mid + 1;
       }
         
       // If there is a better solution
       // then update the answer
       if (ans < a)
           a = ans;
    }
      
    // Condition if the solution 
    // doesn't exist in the matrix
    if (a == int.MaxValue)
        return -1;
    return a + 1;
}
  
// Driver Code
public static void Main(String[] args)
{
    int [,]mat = { { 0, 0, 0 },
                   { 0, 0, 1 },
                   { 0, 1, 1 } };
                      
    Console.Write(search(mat, 3, 3));
}
}
  
// This code is contributed by Amit Katiyar

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Output:

2

Performance Analysis:

  • Time Complexity: O(N*logN)
  • Auxiliary Space: O(1)

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