// 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); }