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// 2643. Row With Maximum Ones
#include "leetcode.h"
/*
* given an 'm * 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 number should be selected. return an array containing the
* index of the row and the number of ones in it
*/
int *rowAndMaximumOnes(int **mat, int matSize, int *matColSize,
int *returnSize) {
int *ans = (int *)malloc(2 * sizeof(int));
int sum = 0, k = 0;
for (int i = 0; i < matSize; i++) {
int tmp = 0;
for (int j = 0; j < *matColSize; j++)
if (mat[i][j])
tmp++;
if (sum < tmp) {
sum = tmp;
k = i;
}
}
ans[0] = k;
ans[1] = sum;
*returnSize = 2;
return ans;
}
int main() {
int m1[2][2] = {{0, 1}, {1, 0}}, m2[2][3] = {{0, 0, 0}, {0, 1, 1}},
m3[3][2] = {{0, 0}, {1, 1}, {0, 0}}, mcs1 = 2, mcs2 = 2, mcs3 = 3, rs1,
rs2, rs3;
int *ramo1 = rowAndMaximumOnes((int **)m1, ARRAY_SIZE(m1), &mcs1, &rs1);
int *ramo2 = rowAndMaximumOnes((int **)m2, ARRAY_SIZE(m2), &mcs2, &rs2);
int *ramo3 = rowAndMaximumOnes((int **)m3, ARRAY_SIZE(m3), &mcs3, &rs3);
for (int i = 0; i < rs1; i++)
printf("%d\n", ramo1[i]); // expect: 0 1
printf("\n");
for (int i = 0; i < rs2; i++)
printf("%d\n", ramo1[i]); // expect: 1 2
printf("\n");
for (int i = 0; i < rs3; i++)
printf("%d\n", ramo1[i]); // expect: 1 2
printf("\n");
free(ramo1), free(ramo2), free(ramo3);
}
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