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-rw-r--r--row_max_ones.c49
1 files changed, 49 insertions, 0 deletions
diff --git a/row_max_ones.c b/row_max_ones.c
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+++ b/row_max_ones.c
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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);
+}