<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]]
What would be the output of the following call :
int len1, len2;
int **B = performOps(A, m, n, &len1, &len2);
int i, j;
for (i = 0; i < len1; i++) {
for (j = 0; j < len2; j++) {
printf("%d ", B[i][j]);
}
}
Answer:-
Answer is := 4 3 2 1 8 7 6 5 12 11 10 9
There is no shortcut to solve this problem keep all step in mind.
Myprogworld 