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// 1920. Build Array from Permutation
#include "leetcode.h"
/*
* given a 0-based permutation 'nums', build an array 'ans' of the same length
* where 'ans[i] = nums[nums[i]]' for each '0 <= i < nums.len' and return it. a
* 0-based permutation 'nums' is an array of distinct integers from 0 to
* 'nums.len - 1'.
*/
int *buildArray(int *nums, int numsSize, int *returnSize) {
*returnSize = numsSize;
for (int i = 0; i < numsSize; i++)
nums[i] += (nums[nums[i]] & 1023) << 10;
for (int i = 0; i < numsSize; i++)
nums[i] >>= 10;
return nums;
}
int main() {
int n1[] = {0, 2, 1, 5, 3, 4}, rs1;
int n2[] = {5, 0, 1, 2, 3, 4}, rs2;
int *ba1 = buildArray(n1, ARRAY_SIZE(n1), &rs1);
int *ba2 = buildArray(n2, ARRAY_SIZE(n2), &rs2);
for (int i = 0; i < rs1; i++)
printf("%d ", ba1[i]); // expect: 0 1 2 4 5 3
printf("\n");
for (int i = 0; i < rs2; i++)
printf("%d ", ba2[i]); // expect: 4 5 0 1 2 3
printf("\n");
}
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