// 3022. Minimize OR of Remaining Elements Using Operations #include "leetcode.h" /* * given a 0 indexed integer array 'nums' and an integer 'k'. in one operation, * you can pick any index 'i' of 'nums' such that '0 <= i < nums.len - 1' and * replace 'nums[i]' and 'nums[i + 1]' with a single occurence of 'nums[i] & * nums[i + 1]' where & represents the bitwise and operation. return the minimum * possible value of the bitwise or of the remaining elements of 'nums' after * applying at most 'k' operations. */ int minOrAfterOperations(int *nums, int numsSize, int k) { int max_elem = nums[0], v = (1 << 30) - 1, mask = 0; for (int i = 1; i < numsSize; i++) if (nums[i] > max_elem) max_elem = nums[i]; for (int i = 29; i >= 0; --i) { int curr = 1 << i; mask |= curr; if (curr > max_elem) continue; int tmp = numsSize; bool vis = false; for (int j = 0; j < numsSize && tmp > k; --tmp) { int c = nums[i++]; for (; i < tmp && c & mask; c &= nums[i++]) ; if (c & mask) break; vis = true; } if (!vis || tmp > k) mask ^= curr; } return v ^ mask; } int main() { int n1[] = {3, 5, 3, 2, 7}, n2[] = {7, 3, 15, 14, 2, 8}, n3[] = {10, 7, 10, 3, 9, 14, 9, 4}; printf("%d\n", minOrAfterOperations(n1, ARRAY_SIZE(n1), 2)); // expect: 3 printf("%d\n", minOrAfterOperations(n2, ARRAY_SIZE(n2), 4)); // expect: 2 printf("%d\n", minOrAfterOperations(n3, ARRAY_SIZE(n3), 15)); // expect: 15 }