# Find the row with maximum number of 1s

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
```

## Recommended: Please solve it on “PRACTICE ” first, before moving on to the solution.

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 max. Finally, return the index of row with maximum 1s. The time complexity of this method is O(m*n) where m is number of rows and n is number of columns in matrix.

We can do better. Since each row is sorted, we can use Binary Search to count of 1s in each row. We find the index of first instance of 1 in each row. The count of 1s will be equal to total number of columns minus the index of first 1.

See the following code for implementation of the above approach.

## C++

 `// CPP program to find the row  ` `// with maximum number of 1s  ` `#include ` `using` `namespace` `std; ` `#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 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 ` `#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 Code ` `int` `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 1s ` `import` `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'. `

## Python 3

 `# 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 Code ` `mat ``=` `[[``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 `

## PHP

 `= ``\$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(mLogn) where m is number of rows and n is number of columns in matrix.

The above solution can be optimized further. Instead of doing 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 binary search in complete row, we do search in before the index of last max.

Following is an optimized version of the above solution.

## C++

 `#include ` `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, 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, 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; ` `} `

The worst case time complexity of the above optimized version is also O(mLogn), the will solution work better on average. 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 1

Following 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.

Following is C++ implementation of this method.

 `// 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` `max_row_index = 0; ` ` `  `    ``// The function first() returns index of first 1 in row 0. ` `    ``// Use this index to initialize the index of leftmost 1 seen so far ` `    ``int` `j = first(mat, 0, C-1); ` `    ``if` `(j == -1) ``// if 1 is not present in first row ` `      ``j = C - 1; ` ` `  `    ``for` `(``int` `i = 1; 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 ` `        ``} ` `    ``} ` `    ``return` `max_row_index; ` `} `

Thanks to Tylor, Ankan and Palash for their inputs.

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Improved By : shrikanth13, rathbhupendra

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