Given a boolean 2D array, where each row is sorted. Find the row with the maximum number of 1s.
Example:
Input matrix : 0 1 1 1
0 0 1 1
1 1 1 1 // this row has maximum 1s
0 0 0 0
Output: 2
A simple method is to do a row-wise traversal of the matrix, count the number of 1s in each row, and compare the count with the max. Finally, return the index of the row with a maximum of 1s. The time complexity of this method is O(m*n) where m is the number of rows and n is the number of columns in the matrix.
Implementation:
C++
// CPP program to find the row // with maximum number of 1s #include <bits/stdc++.h>using namespace std;#define R 4 #define C 4 // Function that returns index of row // with maximum number of 1s. int rowWithMax1s(bool mat[R][C]) { // code here int rowIndex = -1 ; int maxCount = 0 ; for(int i = 0 ; i < R ; i++){ int count = 0 ; for(int j = 0 ; j < C ; j++ ){ if(mat[i][j] == 1){ count++ ; } } if(count > maxCount){ maxCount = count ; rowIndex = i ; } } return rowIndex ;}// Driver Code int main() { bool mat[R][C] = { {0, 0, 0, 1}, {0, 1, 1, 1}, {1, 1, 1, 1}, {0, 0, 0, 0}}; cout << "Index of row with maximum 1s is " << rowWithMax1s(mat); return 0; } |
C
// C program to find the row with maximum number of 1s.#include<stdio.h>#include<stdbool.h> #define R 4#define C 4// Function that returns index of row// with maximum number of 1s.int rowWithMax1s(bool mat[R][C]) { int indexOfRowWithMax1s = -1 ; int maxCount = 0 ; // Visit each row. // Count number of 1s. /* If count is more that the maxCount then update the maxCount and store the index of current row in indexOfRowWithMax1s variable. */ for(int i = 0 ; i < R ; i++){ int count = 0 ; for(int j = 0 ; j < C ; j++ ){ if(mat[i][j] == 1){ count++ ; } } if(count > maxCount){ maxCount = count ; indexOfRowWithMax1s = i ; } } return indexOfRowWithMax1s ;} // Driver Codeint main(){ bool mat[R][C] = { {0, 0, 0, 1}, {0, 1, 1, 1}, {1, 1, 1, 1}, {0, 0, 0, 0}}; int indexOfRowWithMax1s = rowWithMax1s(mat); printf("Index of row with maximum 1s is %d",indexOfRowWithMax1s); return 0;}// This code is contributed by Rohit_Dwivedi |
Java
// Java program for the above approachimport java.util.*;class GFG {static int R = 4 ;static int C = 4 ;// Function that returns index of row// with maximum number of 1s.static int rowWithMax1s(int mat[][], int R, int C){ // Flag to check if there is not even a single 1 in the matrix. boolean flag = true; // Initialize max values int max_row_index = 0, max_ones = 0;; // Traverse for each row and count number of 1s for(int i = 0 ; i < R ; i++){ int count1 = 0 ; for(int j = 0 ; j < C ; j++){ if(mat[i][j] == 1){ count1++; flag = false; } } if(count1>max_ones){ max_ones = count1; max_row_index = i; } } // Edge case where there are no 1 in the matrix if(flag){ return -1; } return max_row_index;} // Driver Code public static void main(String[] args) { int mat[][] = { {0, 0, 0, 1}, {0, 1, 1, 1}, {1, 1, 1, 1}, {0, 0, 0, 0}}; System.out.print("Index of row with maximum 1s is " + rowWithMax1s(mat,R,C)); }} |
Python3
# Python implementation of the approachR,C = 4,4# Function to find the index of first index # of 1 in a boolean array arr def first(arr , low , high): if(high >= low): # Get the middle index mid = low + (high - low)//2 # Check if the element at middle index is first 1 if ( ( mid == 0 or arr[mid-1] == 0) and arr[mid] == 1): return mid # If the element is 0, recur for right side elif (arr[mid] == 0): return first(arr, (mid + 1), high); # If element is not first 1, recur for left side else: return first(arr, low, (mid -1)); return -1# Function that returns index of row # with maximum number of 1s. def rowWithMax1s(mat): # Initialize max values max_row_index,Max = 0,-1 # Traverse for each row and count number of 1s # by finding the index of first 1 for i in range(R): index = first (mat[i], 0, C-1) if (index != -1 and C-index > Max): Max = C - index; max_row_index = i return max_row_index # Driver Codemat = [[0, 0, 0, 1], [0, 1, 1, 1], [1, 1, 1, 1], [0, 0, 0, 0]]print("Index of row with maximum 1s is " + str(rowWithMax1s(mat)))# This code is contributed by shinjanpatra |
C#
// C# program for the above approachusing System;using System.Collections.Generic;public class GFG { static int R = 4; static int C = 4; // Function to find the index of first index // of 1 in a bool array []arr static int first(int []arr, int low, int high) { if (high >= low) { // Get the middle index int mid = low + (high - low) / 2; // Check if the element at middle index is first 1 if ((mid == 0 || arr[mid - 1] == 0) && arr[mid] == 1) return mid; // If the element is 0, recur for right side else if (arr[mid] == 0) return first(arr, (mid + 1), high); // If element is not first 1, recur for left side else return first(arr, low, (mid - 1)); } return -1; } public static int[] GetRow(int[,] matrix, int row) { var rowLength = matrix.GetLength(1); var rowVector = new int[rowLength]; for (var i = 0; i < rowLength; i++) rowVector[i] = matrix[row, i]; return rowVector; } // Function that returns index of row // with maximum number of 1s. static int rowWithMax1s(int [,]mat) { // Initialize max values int max_row_index = 0, max = -1; // Traverse for each row and count number of 1s // by finding the index of first 1 int i, index; for (i = 0; i < R; i++) { int []row = GetRow(mat,i); index = first(row, 0, C - 1); if (index != -1 && C - index > max) { max = C - index; max_row_index = i; } } return max_row_index; } // Driver Code public static void Main(String[] args) { int [,]mat = { { 0, 0, 0, 1 }, { 0, 1, 1, 1 }, { 1, 1, 1, 1 }, { 0, 0, 0, 0 } }; Console.Write("Index of row with maximum 1s is " + rowWithMax1s(mat)); }}// This code is contributed by Rajput-Ji |
Javascript
<script>// javascript program for the above approachvar R = 4 ;var C = 4 ;// Function to find the index of first index // of 1 in a boolean array arr function first(arr , low , high) { if(high >= low) { // Get the middle index var mid = low + parseInt((high - low)/2); // Check if the element at middle index is first 1 if ( ( mid == 0 || arr[mid-1] == 0) && arr[mid] == 1) return mid; // If the element is 0, recur for right side else if (arr[mid] == 0) return first(arr, (mid + 1), high); // If element is not first 1, recur for left side else return first(arr, low, (mid -1)); } return -1; } // Function that returns index of row // with maximum number of 1s. function rowWithMax1s(mat) { // Initialize max values var max_row_index = 0, max = -1; // Traverse for each row and count number of 1s // by finding the index of first 1 var i, index; for (i = 0; i < R; i++) { index = first (mat[i], 0, C-1); if (index != -1 && C-index > max) { max = C - index; max_row_index = i; } } return max_row_index; } // Driver Code var mat = [[0, 0, 0, 1], [0, 1, 1, 1], [1, 1, 1, 1], [0, 0, 0, 0]]; document.write("Index of row with maximum 1s is " + rowWithMax1s(mat)); // This code is contributed by Rajput-Ji</script> |
Output
Index of row with maximum 1s is 2
Time Complexity: O(m*n)
Auxiliary Space: O(1)
We can do better. Since each row is sorted, we can use Binary Search to count 1s in each row. We find the index of the first instance of 1 in each row. The count of 1s will be equal to the total number of columns minus the index of the first 1.
Implementation: See the following code for the implementation of the above approach.
C++
// CPP program to find the row // with maximum number of 1s #include <bits/stdc++.h>using namespace std;#define R 4 #define C 4 // Function to find the index of first instance // of 1 in a boolean array arr[] int first(bool arr[], int low, int high) { if(high >= low) { // Get the middle index int mid = low + (high - low)/2; // Check if the element at middle index is first 1 if ( ( mid == 0 || arr[mid-1] == 0) && arr[mid] == 1) return mid; // If the element is 0, recur for right side else if (arr[mid] == 0) return first(arr, (mid + 1), high); // If element is not first 1, recur for left side else return first(arr, low, (mid -1)); } return -1; } // Function that returns index of row // with maximum number of 1s. int rowWithMax1s(bool mat[R][C]) { // Initialize max values int max_row_index = 0, max = -1; // Traverse for each row and count number of 1s // by finding the index of first 1 int i, index; for (i = 0; i < R; i++) { index = first (mat[i], 0, C-1); if (index != -1 && C-index > max) { max = C - index; max_row_index = i; } } return max_row_index; } // Driver Code int main() { bool mat[R][C] = { {0, 0, 0, 1}, {0, 1, 1, 1}, {1, 1, 1, 1}, {0, 0, 0, 0}}; cout << "Index of row with maximum 1s is " << rowWithMax1s(mat); return 0; } // This is code is contributed by rathbhupendra |
C
// C program to find the row// with maximum number of 1s#include <stdio.h>#define R 4#define C 4// Function to find the index of first index// of 1 in a boolean array arr[] int first(bool arr[], int low, int high){if(high >= low){ // Get the middle index int mid = low + (high - low)/2; // Check if the element at middle index is first 1 if ( ( mid == 0 || arr[mid-1] == 0) && arr[mid] == 1) return mid; // If the element is 0, recur for right side else if (arr[mid] == 0) return first(arr, (mid + 1), high); // If element is not first 1, recur for left side else return first(arr, low, (mid -1));}return -1;}// Function that returns index of row// with maximum number of 1s. int rowWithMax1s(bool mat[R][C]){ // Initialize max values int max_row_index = 0, max = -1; // Traverse for each row and count number of 1s // by finding the index of first 1 int i, index; for (i = 0; i < R; i++) { index = first (mat[i], 0, C-1); if (index != -1 && C-index > max) { max = C - index; max_row_index = i; } } return max_row_index;}// Driver Codeint main(){ bool mat[R][C] = { {0, 0, 0, 1}, {0, 1, 1, 1}, {1, 1, 1, 1}, {0, 0, 0, 0}}; printf("Index of row with maximum 1s is %d " , rowWithMax1s(mat)); return 0;} |
Java
// Java program to find the row// with maximum number of 1simport java.io.*;class GFG { static int R = 4, C = 4; // Function to find the index of first index // of 1 in a boolean array arr[] static int first(int arr[], int low, int high) { if (high >= low) { // Get the middle index int mid = low + (high - low) / 2; // Check if the element at middle index is first 1 if ((mid == 0 || (arr[mid - 1] == 0)) && arr[mid] == 1) return mid; // If the element is 0, recur for right side else if (arr[mid] == 0) return first(arr, (mid + 1), high); // If element is not first 1, recur for left side else return first(arr, low, (mid - 1)); } return -1; } // Function that returns index of row // with maximum number of 1s. static int rowWithMax1s(int mat[][]) { // Initialize max values int max_row_index = 0, max = -1; // Traverse for each row and count number of // 1s by finding the index of first 1 int i, index; for (i = 0; i < R; i++) { index = first(mat[i], 0, C - 1); if (index != -1 && C - index > max) { max = C - index; max_row_index = i; } } return max_row_index; } // Driver Code public static void main(String[] args) { int mat[][] = { { 0, 0, 0, 1 }, { 0, 1, 1, 1 }, { 1, 1, 1, 1 }, { 0, 0, 0, 0 } }; System.out.println("Index of row with maximum 1s is " + rowWithMax1s(mat)); }}// This code is contributed by 'Gitanjali'. |
Python3
# Python3 program to find the row# with maximum number of 1s# Function to find the index# of first index of 1 in a# boolean array arr[]def first(arr, low, high): if high >= low: # Get the middle index mid = low + (high - low)//2 # Check if the element at # middle index is first 1 if (mid == 0 or arr[mid - 1] == 0) and arr[mid] == 1: return mid # If the element is 0, # recur for right side elif arr[mid] == 0: return first(arr, (mid + 1), high) # If element is not first 1, # recur for left side else: return first(arr, low, (mid - 1)) return -1# Function that returns# index of row with maximum# number of 1s.def rowWithMax1s(mat): # Initialize max values R = len(mat) C = len(mat[0]) max_row_index = 0 max = -1 # Traverse for each row and # count number of 1s by finding # the index of first 1 for i in range(0, R): index = first(mat[i], 0, C - 1) if index != -1 and C - index > max: max = C - index max_row_index = i return max_row_index# Driver Codemat = [[0, 0, 0, 1], [0, 1, 1, 1], [1, 1, 1, 1], [0, 0, 0, 0]]print("Index of row with maximum 1s is", rowWithMax1s(mat))# This code is contributed# by shreyanshi_arun |
C#
// C# program to find the row with maximum// number of 1s using System;class GFG{public static int R = 4, C = 4;// Function to find the index of first index // of 1 in a boolean array arr[] public static int first(int[] arr, int low, int high){ if (high >= low) { // Get the middle index int mid = low + (high - low) / 2; // Check if the element at middle // index is first 1 if ((mid == 0 || (arr[mid - 1] == 0)) && arr[mid] == 1) { return mid; } // If the element is 0, recur // for right side else if (arr[mid] == 0) { return first(arr, (mid + 1), high); } // If element is not first 1, recur // for left side else { return first(arr, low, (mid - 1)); } } return -1;}// Function that returns index of row // with maximum number of 1s. public static int rowWithMax1s(int[][] mat){ // Initialize max values int max_row_index = 0, max = -1; // Traverse for each row and count number // of 1s by finding the index of first 1 int i, index; for (i = 0; i < R; i++) { index = first(mat[i], 0, C - 1); if (index != -1 && C - index > max) { max = C - index; max_row_index = i; } } return max_row_index;}// Driver Code public static void Main(string[] args){ int[][] mat = new int[][] { new int[] {0, 0, 0, 1}, new int[] {0, 1, 1, 1}, new int[] {1, 1, 1, 1}, new int[] {0, 0, 0, 0} }; Console.WriteLine("Index of row with maximum 1s is " + rowWithMax1s(mat));}}// This code is contributed by Shrikant13 |
Javascript
<script> // JavaScript program to find the row // with maximum number of 1s R = 4 C = 4 // Function to find the index of first index // of 1 in a boolean array arr[] const first = (arr, low, high) => { if (high >= low) { // Get the middle index let mid = low + parseInt((high - low) / 2); // Check if the element at middle index is first 1 if ((mid == 0 || arr[mid - 1] == 0) && arr[mid] == 1) return mid; // If the element is 0, recur for right side else if (arr[mid] == 0) return first(arr, (mid + 1), high); // If element is not first 1, recur for left side else return first(arr, low, (mid - 1)); } return -1; } // Function that returns index of row // with maximum number of 1s. const rowWithMax1s = (mat) => { // Initialize max values let max_row_index = 0, max = -1; // Traverse for each row and count number of 1s // by finding the index of first 1 let i, index; for (i = 0; i < R; i++) { index = first(mat[i], 0, C - 1); if (index != -1 && C - index > max) { max = C - index; max_row_index = i; } } return max_row_index; } // Driver Code let mat = [ [0, 0, 0, 1], [0, 1, 1, 1], [1, 1, 1, 1], [0, 0, 0, 0] ]; document.write(`Index of row with maximum 1s is ${rowWithMax1s(mat)}`); // This code is contributed by rakeshsahni</script> |
PHP
<?php// PHP program to find the row // with maximum number of 1s define("R", 4);define("C", 4); // Function to find the index of first // index of 1 in a boolean array arr[] function first($arr, $low, $high) { if($high >= $low) { // Get the middle index $mid = $low + intval(($high - $low) / 2); // Check if the element at middle // index is first 1 if (($mid == 0 || $arr[$mid - 1] == 0) && $arr[$mid] == 1) return $mid; // If the element is 0, recur for // right side else if ($arr[$mid] == 0) return first($arr, ($mid + 1), $high); // If element is not first 1, recur // for left side else return first($arr, $low, ($mid - 1)); } return -1; } // Function that returns index of row // with maximum number of 1s. function rowWithMax1s($mat) { // Initialize max values $max_row_index = 0; $max = -1; // Traverse for each row and count number // of 1s by finding the index of first 1 for ($i = 0; $i < R; $i++) { $index = first ($mat[$i], 0, (C - 1)); if ($index != -1 && (C - $index) > $max) { $max = C - $index; $max_row_index = $i; } } return $max_row_index; } // Driver Code $mat = array(array(0, 0, 0, 1), array(0, 1, 1, 1), array(1, 1, 1, 1), array(0, 0, 0, 0)); echo "Index of row with maximum 1s is " . rowWithMax1s($mat); // This code is contributed by rathbhupendra?> |
Output
Index of row with maximum 1s is 2
Time Complexity: O(m log n) where m is the number of rows and n is the number of columns in the matrix.
Auxiliary Space: O(log n), as implicit stack is created due to recursion.
The above solution can be optimized further. Instead of doing a binary search in every row, we first check whether the row has more 1s than max so far. If the row has more 1s, then only count 1s in the row. Also, to count 1s in a row, we don’t do a binary search in a complete row, we do a search before the index of the last max.
Implementation: Following is an optimized version of the above solution.
C++
#include <bits/stdc++.h>using namespace std;// The main function that returns index// of row with maximum number of 1s. int rowWithMax1s(bool mat[R][C]) { int i, index; // Initialize max using values from first row. int max_row_index = 0; int max = first(mat[0], 0, C - 1); // Traverse for each row and count number of 1s // by finding the index of first 1 for (i = 1; i < R; i++) { // Count 1s in this row only if this row // has more 1s than max so far // Count 1s in this row only if this row // has more 1s than max so far if (max != -1 && mat[i][C - max - 1] == 1) { // Note the optimization here also index = first (mat[i], 0, C - max); if (index != -1 && C - index > max) { max = C - index; max_row_index = i; } } else { max = first(mat[i], 0, C - 1); } } return max_row_index; } // This code is contributed by rathbhupendra |
C
// The main function that returns index of row with maximum number of 1s.int rowWithMax1s(bool mat[R][C]){ int i, index; // Initialize max using values from first row. int max_row_index = 0; int max = first(mat[0], 0, C-1); // Traverse for each row and count number of 1s by finding the index // of first 1 for (i = 1; i < R; i++) { // Count 1s in this row only if this row has more 1s than // max so far // Count 1s in this row only if this row has more 1s than // max so far if (max != -1 && mat[i][C-max-1] == 1) { // Note the optimization here also index = first (mat[i], 0, C-max); if (index != -1 && C-index > max) { max = C - index; max_row_index = i; } } else { max = first(mat[i], 0, C - 1); } } return max_row_index;} |
Java
public class gfg{ // The main function that returns index // of row with maximum number of 1s. static int rowWithMax1s(boolean mat[][]) { int i, index; // Initialize max using values from first row. int max_row_index = 0; int max = first(mat[0], 0, C - 1); // Traverse for each row and count number of 1s // by finding the index of first 1 for (i = 1; i < R; i++) { // Count 1s in this row only if this row // has more 1s than max so far // Count 1s in this row only if this row // has more 1s than max so far if (max != -1 && mat[i][C - max - 1] == 1) { // Note the optimization here also index = first (mat[i], 0, C - max); if (index != -1 && C - index > max) { max = C - index; max_row_index = i; } } else { max = first(mat[i], 0, C - 1); } } return max_row_index; }}// This code is contributed by divyesh072019. |
Python3
# The main function that returns index# of row with maximum number of 1s.def rowWithMax1s(mat): # Initialize max using values from first row. max_row_index = 0 max = first(mat[0], 0, C - 1) # Traverse for each row and count number of 1s # by finding the index of first 1 for i in range(1, R): # Count 1s in this row only if this row # has more 1s than max so far # Count 1s in this row only if this row # has more 1s than max so far if (max != -1 and mat[i][C - max - 1] == 1): # Note the optimization here also index = first(mat[i], 0, C - max) if (index != -1 and C - index > max): max = C - index max_row_index = i else: max = first(mat[i], 0, C - 1) return max_row_index# This code is contributed by Dharanendra L V |
C#
using System;class GFG { // The main function that returns index // of row with maximum number of 1s. static int rowWithMax1s(bool[,] mat) { int i, index; // Initialize max using values from first row. int max_row_index = 0; int max = first(mat[0], 0, C - 1); // Traverse for each row and count number of 1s // by finding the index of first 1 for (i = 1; i < R; i++) { // Count 1s in this row only if this row // has more 1s than max so far // Count 1s in this row only if this row // has more 1s than max so far if (max != -1 && mat[i,C - max - 1] == 1) { // Note the optimization here also index = first (mat[i], 0, C - max); if (index != -1 && C - index > max) { max = C - index; max_row_index = i; } } else { max = first(mat[i], 0, C - 1); } } return max_row_index; }}// This code is contributed by divyeshrabadiya07. |
Javascript
<script> // The main function that returns index// of row with maximum number of 1s.function rowWithMax1s(mat){ let i, index; // Initialize max using values from first row. let max_row_index = 0; let max = first(mat[0], 0, C - 1); // Traverse for each row and count number of 1s // by finding the index of first 1 for(i = 1; i < R; i++) { // Count 1s in this row only if this row // has more 1s than max so far // Count 1s in this row only if this row // has more 1s than max so far if (max != -1 && mat[i][C - max - 1] == 1) { // Note the optimization here also index = first (mat[i], 0, C - max); if (index != -1 && C - index > max) { max = C - index; max_row_index = i; } } else { max = first(mat[i], 0, C - 1); } } return max_row_index;}// This code is contributed by suresh07</script> |
Time complexity: O(m log n),
Auxiliary Space: O(log n)
Thanks to Naveen Kumar Singh for suggesting the above solution.
The worst case of the above solution occurs for a matrix like following.
0 0 0 … 0 1
0 0 0 ..0 1 1
0 … 0 1 1 1
….0 1 1 1 1Following method works in O(m+n) time complexity in worst case.
- Step1: Get the index of first (or leftmost) 1 in the first row.
- Step2: Do following for every row after the first row
- …IF the element on left of previous leftmost 1 is 0, ignore this row.
- …ELSE Move left until a 0 is found. Update the leftmost index to this index and max_row_index to be the current row.
- The time complexity is O(m+n) because we can possibly go as far left as we came ahead in the first step.
Implementation: Following is the implementation of this method.
C++
// C++ program to find the row with maximum// number of 1s#include <bits/stdc++.h>using namespace std;#define R 4#define C 4// The main function that returns index of row with maximum// number of 1s.int rowWithMax1s(bool mat[R][C]){ // Initialize first row as row with max 1s int j,max_row_index = 0; j = C - 1; for (int i = 0; i < R; i++) { // Move left until a 0 is found bool flag=false; //to check whether a row has more 1's than previous while (j >= 0 && mat[i][j] == 1) { j = j - 1; // Update the index of leftmost 1 // seen so far flag=true ;//present row has more 1's than previous } // if the present row has more 1's than previous if(flag){ max_row_index = i; // Update max_row_index } } if(max_row_index==0&&mat[0][C-1]==0) return -1; return max_row_index;}// Driver Codeint main(){ bool mat[R][C] = { {0, 0, 0, 1}, {0, 1, 1, 1}, {1, 1, 1, 1}, {0, 0, 0, 0}}; cout << "Index of row with maximum 1s is " << rowWithMax1s(mat); return 0;}// this code is contributed by Rishabh Chauhan |
Java
// Java program to find the row// with maximum number of 1simport java.io.*;class GFG { static int R = 4, C = 4; // Function that returns index of row // with maximum number of 1s. static int rowWithMax1s(int mat[][]) { // Initialize first row as row with max 1s int j,max_row_index = 0; j = C - 1; for (int i = 0; i < R; i++) { // Move left until a 0 is found while (j >= 0 && mat[i][j] == 1) { j = j - 1; // Update the index of leftmost 1 // seen so far max_row_index = i; // Update max_row_index } } if(max_row_index==0&&mat[0][C-1]==0) return -1; return max_row_index; } // Driver Code public static void main(String[] args) { int mat[][] = { { 0, 0, 0, 1 }, { 0, 1, 1, 1 }, { 1, 1, 1, 1 }, { 0, 0, 0, 0 } }; System.out.println("Index of row with maximum 1s is "+ rowWithMax1s(mat)); }}// This code is contributed by 'Rishabh Chauhan'. |
Python3
# Python3 program to find the row# with maximum number of 1s# Function that returns# index of row with maximum# number of 1s.def rowWithMax1s(mat): # Initialize max values R = len(mat) C = len(mat[0]) max_row_index = 0 index = C-1 # Traverse for each row and # count number of 1s by finding # the index of first 1 for i in range(0, R): flag = False # to check whether a row has more 1's than previous while(index >= 0 and mat[i][index] == 1): flag = True # present row has more 1's than previous index -= 1 if(flag): # if the present row has more 1's than previous max_row_index = i if max_row_index == 0 and mat[0][C-1] == 0: return 0 return max_row_index# Driver Codemat = [[0, 0, 0, 1], [0, 1, 1, 1], [1, 1, 1, 1], [0, 0, 0, 0]]print("Index of row with maximum 1s is", rowWithMax1s(mat))# This code is contributed# by Rishabh Chauhan |
C#
// C# program to find the row with maximum// number of 1susing System;class GFG{public static int R = 4, C = 4;// Function that returns index of row// with maximum number of 1s.public static int rowWithMax1s(int[][] mat){ // Initialize max values int max_row_index = 0; int i, index; index=C-1; for (i = 0; i < R; i++) { if (index >=0 && mat[i][index]==1) { index-=1; max_row_index = i; } } if (max_row_index==0&&mat[0][C-1]==0) return 0; return max_row_index;}// Driver Codepublic static void Main(string[] args){ int[][] mat = new int[][] { new int[] {0, 0, 0, 1}, new int[] {0, 1, 1, 1}, new int[] {1, 1, 1, 1}, new int[] {0, 0, 0, 0} }; Console.WriteLine("Index of row with maximum 1s is " +rowWithMax1s(mat));}}// This code is contributed by Rishabh Chauhan |
Javascript
<script> // JavaScript program for the above approach let R = 4 let C = 4 // The main function that returns index of row with maximum // number of 1s. function rowWithMax1s(mat) { // Initialize first row as row with max 1s let j, max_row_index = 0; j = C - 1; for (let i = 0; i < R; i++) { // Move left until a 0 is found let flag = false; // to check whether a row has more 1's than previous while (j >= 0 && mat[i][j] == 1) { j = j - 1; // Update the index of leftmost 1 // seen so far flag = true;//present row has more 1's than previous } // if the present row has more 1's than previous if (flag) { max_row_index = i; // Update max_row_index } } if (max_row_index == 0 && mat[0][C - 1] == 0) return -1; return max_row_index; } // Driver Code let mat = [[0, 0, 0, 1], [0, 1, 1, 1], [1, 1, 1, 1], [0, 0, 0, 0]]; document.write("Index of row with maximum 1s is " + rowWithMax1s(mat));// This code is contributed by Potta Lokesh </script> |
Output
Index of row with maximum 1s is 2
Time Complexity: O(m+n) where m is the number of rows and n is the number of columns in the matrix.
Auxiliary Space: O(1)
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