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-rw-r--r-- | row_max_ones.c | 49 | ||||
-rw-r--r-- | row_max_ones.py | 36 |
2 files changed, 85 insertions, 0 deletions
diff --git a/row_max_ones.c b/row_max_ones.c new file mode 100644 index 0000000..9a18000 --- /dev/null +++ b/row_max_ones.c @@ -0,0 +1,49 @@ +// 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); +} diff --git a/row_max_ones.py b/row_max_ones.py new file mode 100644 index 0000000..2b0004e --- /dev/null +++ b/row_max_ones.py @@ -0,0 +1,36 @@ +# 2643. Row With Maximum Ones + +""" +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 +""" + + +class Solution(object): + def rowAndMaximumOnes(self, mat): + """ + :type mat: List[List[int]] + :rtype: List[int] + """ + cnt, ans = 0, [] + ans.append(0) + ans.append(0) + for row in range(len(mat)): + for col in mat[row]: + if col == 1: + cnt += 1 + if ans[1] < cnt: + ans[0] = row + ans[1] = cnt + cnt = 0 + return ans + + +if __name__ == "__main__": + obj = Solution() + print(obj.rowAndMaximumOnes(mat=[[0, 1], [1, 0]])) + print(obj.rowAndMaximumOnes(mat=[[0, 0, 0], [0, 1, 1]])) + print(obj.rowAndMaximumOnes(mat=[[0, 0], [1, 1], [0, 0]])) |