#arrayt

<br />int** performOps(int **A, int m, int n, int *len1, int *len2) { int i, j; *len1 = m; *len2 = n; int **B = (int **)malloc((*len1) * sizeof(int *)); for (i = 0; i < *len1; i++) { B[i] = (int *)malloc((*len2) * sizeof(int)); } for (i = 0; i < m; i++) { for (j = 0; j < n; j++) { B[i][n - 1 - j] = A[i][j]; } } return B; }

Let’s say m=3, n=4, and A : [[1,2,3,4], [5,6,7,8], [9,10, 11, 12]]

Wha…