// 1829. Maximum XOR for Each Query #include "leetcode.h" /* * you are given a sorted array 'nums' of 'n' non negative integers and an * integer 'maximumBit'. you want to perform the following query 'n' times: find * a non-negative integer 'k < 2^maximumBit' such that 'nums[0] ^ nums[1] ^ ... * ^ nums[n]' is maximised. 'k' is the answer to the i'th query. remove the last * element from the current array 'nums'. return an array 'answer' where * 'answer[i]' is the answer to the i'th query. */ int *getMaximumXor(int *nums, int numsSize, int maximumBit, int *returnSize) { int *ans = (int *)malloc(numsSize * sizeof(int)); *returnSize = numsSize; int max = pow(2, maximumBit) - 1; for (int i = 1; i < numsSize; i++) nums[i] ^= nums[i - 1]; for (int i = 0; i < numsSize; i++) ans[i] = nums[numsSize - i - 1] ^ max; return ans; } int main() { int n1[] = {0, 1, 1, 3}, rs1; int n2[] = {2, 3, 4, 7}, rs2; int n3[] = {0, 1, 2, 2, 5, 7}, rs3; int *gmx1 = getMaximumXor(n1, ARRAY_SIZE(n1), 2, &rs1); int *gmx2 = getMaximumXor(n2, ARRAY_SIZE(n2), 3, &rs2); int *gmx3 = getMaximumXor(n3, ARRAY_SIZE(n3), 3, &rs3); for (int i = 0; i < rs1; i++) printf("%d ", gmx1[i]); // expect: 0 3 2 3 printf("\n"); for (int i = 0; i < rs2; i++) printf("%d ", gmx2[i]); // expect: 5 2 6 5 printf("\n"); for (int i = 0; i < rs3; i++) printf("%d ", gmx3[i]); // expect: 4 3 6 4 6 7 printf("\n"); free(gmx1), free(gmx2), free(gmx3); }