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// 2460. Apply Operations to an Array
#include "leetcode.h"
/*
* you are given 0-indexed array 'nums' of size 'n' consisting of non-negative
* integers. you need to apply 'n - 1' operations to this array where in the
* i'th operation (0-indexed), you will apply the following on the i'th element
* of 'nums': if 'nums[i] == nums[i + 1]', then multiply 'nums[i]' by 2 and set
* 'nums[i + 1]' to operate 0. otherwise, you skip this operation. after
* performing all the operations, shift all the 0's to the end of the array.
* return the resulting array. note that the operations are applied sequentially
* not all at once.
*/
int *applyOperations(int *nums, int numsSize, int *returnSize) {
for (int i = 0; i < numsSize - 1; i++)
if (nums[i] == nums[i + 1] && nums[i]) {
nums[i] *= 2;
nums[i + 1] = 0;
}
int *ans = (int *)malloc(numsSize * sizeof(int)), j = 0;
*returnSize = numsSize;
for (int i = 0; i < numsSize; i++)
if (nums[i])
ans[j++] = nums[i];
while (j < numsSize)
ans[j++] = 0;
return ans;
}
int main() {
int n1[] = {1, 2, 2, 1, 1, 0}, n2[] = {0, 1}, rs1, rs2;
int *ao1 = applyOperations(n1, ARRAY_SIZE(n1), &rs1);
int *ao2 = applyOperations(n2, ARRAY_SIZE(n2), &rs2);
for (int i = 0; i < rs1; i++)
printf("%d ", ao1[i]); // expect: 1 4 2 0 0 0
printf("\n");
for (int i = 0; i < rs2; i++)
printf("%d ", ao2[i]); // expect: 1 0
printf("\n");
free(ao1), free(ao2);
}
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