2643-RowWithMaximumOnes.go

package main

// 2643. Row With Maximum Ones
// Given a m x n binary matrix mat, 
// find the 0-indexed position of the row that contains the maximum count of ones, 
// and the number of ones in that row.

// In case there are multiple rows that have the maximum count of ones, 
// the row with the smallest row number should be selected.

// Return an array containing the index of the row, and the number of ones in it.

// Example 1:
// Input: mat = [[0,1],[1,0]]
// Output: [0,1]
// Explanation: Both rows have the same number of 1's. So we return the index of the smaller row, 0, and the maximum count of ones (1). So, the answer is [0,1]. 

// Example 2:
// Input: mat = [[0,0,0],[0,1,1]]
// Output: [1,2]
// Explanation: The row indexed 1 has the maximum count of ones (2). So we return its index, 1, and the count. So, the answer is [1,2].

// Example 3:
// Input: mat = [[0,0],[1,1],[0,0]]
// Output: [1,2]
// Explanation: The row indexed 1 has the maximum count of ones (2). So the answer is [1,2].

// Constraints:
//     m == mat.length 
//     n == mat[i].length 
//     1 <= m, n <= 100 
//     mat[i][j] is either 0 or 1.

import "fmt"

func rowAndMaximumOnes(mat [][]int) []int {
    mx, row := 0, 0
    for i, cols := range mat {
        count := 0
        for _, v := range cols {
            if v == 1 { count++ }
        }
        if count > mx {
            mx, row = count, i
        }
    }
    return []int{ row, mx }
}

func main() {
    // Example 1:
    // Input: mat = [[0,1],[1,0]]
    // Output: [0,1]
    // Explanation: Both rows have the same number of 1's. So we return the index of the smaller row, 0, and the maximum count of ones (1). So, the answer is [0,1]. 
    fmt.Println(rowAndMaximumOnes([][]int{{0,1},{1,0}})) // [0,1]
    // Example 2:
    // Input: mat = [[0,0,0],[0,1,1]]
    // Output: [1,2]
    // Explanation: The row indexed 1 has the maximum count of ones (2). So we return its index, 1, and the count. So, the answer is [1,2].
    fmt.Println(rowAndMaximumOnes([][]int{{0,0,0},{0,1,1}})) // [1,2]
    // Example 3:
    // Input: mat = [[0,0],[1,1],[0,0]]
    // Output: [1,2]
    // Explanation: The row indexed 1 has the maximum count of ones (2). So the answer is [1,2].
    fmt.Println(rowAndMaximumOnes([][]int{{0,0},{1,1},{0,0}})) // [1,2]
}