Off-by-one on range boundaries
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.
Move from brute-force thinking to an efficient approach using array strategy.
You are given a 0-indexed integer array nums and a positive integer k.
You can apply the following operation on the array any number of times:
0 to 1 or vice versa.Return the minimum number of operations required to make the bitwise XOR of all elements of the final array equal to k.
Note that you can flip leading zero bits in the binary representation of elements. For example, for the number (101)2 you can flip the fourth bit and obtain (1101)2.
Example 1:
Input: nums = [2,1,3,4], k = 1 Output: 2 Explanation: We can do the following operations: - Choose element 2 which is 3 == (011)2, we flip the first bit and we obtain (010)2 == 2. nums becomes [2,1,2,4]. - Choose element 0 which is 2 == (010)2, we flip the third bit and we obtain (110)2 = 6. nums becomes [6,1,2,4]. The XOR of elements of the final array is (6 XOR 1 XOR 2 XOR 4) == 1 == k. It can be shown that we cannot make the XOR equal to k in less than 2 operations.
Example 2:
Input: nums = [2,0,2,0], k = 0 Output: 0 Explanation: The XOR of elements of the array is (2 XOR 0 XOR 2 XOR 0) == 0 == k. So no operation is needed.
Constraints:
1 <= nums.length <= 1050 <= nums[i] <= 1060 <= k <= 106Problem summary: You are given a 0-indexed integer array nums and a positive integer k. You can apply the following operation on the array any number of times: Choose any element of the array and flip a bit in its binary representation. Flipping a bit means changing a 0 to 1 or vice versa. Return the minimum number of operations required to make the bitwise XOR of all elements of the final array equal to k. Note that you can flip leading zero bits in the binary representation of elements. For example, for the number (101)2 you can flip the fourth bit and obtain (1101)2.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Array · Bit Manipulation
[2,1,3,4] 1
[2,0,2,0] 0
minimum-bit-flips-to-convert-number)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2997: Minimum Number of Operations to Make Array XOR Equal to K
class Solution {
public int minOperations(int[] nums, int k) {
for (int x : nums) {
k ^= x;
}
return Integer.bitCount(k);
}
}
// Accepted solution for LeetCode #2997: Minimum Number of Operations to Make Array XOR Equal to K
func minOperations(nums []int, k int) (ans int) {
for _, x := range nums {
k ^= x
}
return bits.OnesCount(uint(k))
}
# Accepted solution for LeetCode #2997: Minimum Number of Operations to Make Array XOR Equal to K
class Solution:
def minOperations(self, nums: List[int], k: int) -> int:
return reduce(xor, nums, k).bit_count()
// Accepted solution for LeetCode #2997: Minimum Number of Operations to Make Array XOR Equal to K
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
pub fn rust_example() {
// Port the logic from the reference block below.
}
}
// Reference (java):
// // Accepted solution for LeetCode #2997: Minimum Number of Operations to Make Array XOR Equal to K
// class Solution {
// public int minOperations(int[] nums, int k) {
// for (int x : nums) {
// k ^= x;
// }
// return Integer.bitCount(k);
// }
// }
// Accepted solution for LeetCode #2997: Minimum Number of Operations to Make Array XOR Equal to K
function minOperations(nums: number[], k: number): number {
for (const x of nums) {
k ^= x;
}
return bitCount(k);
}
function bitCount(i: number): number {
i = i - ((i >>> 1) & 0x55555555);
i = (i & 0x33333333) + ((i >>> 2) & 0x33333333);
i = (i + (i >>> 4)) & 0x0f0f0f0f;
i = i + (i >>> 8);
i = i + (i >>> 16);
return i & 0x3f;
}
Use this to step through a reusable interview workflow for this problem.
Sort the array in O(n log n), then scan for the missing or unique element by comparing adjacent pairs. Sorting requires O(n) auxiliary space (or O(1) with in-place sort but O(n log n) time remains). The sort step dominates.
Bitwise operations (AND, OR, XOR, shifts) are O(1) per operation on fixed-width integers. A single pass through the input with bit operations gives O(n) time. The key insight: XOR of a number with itself is 0, which eliminates duplicates without extra space.
Review these before coding to avoid predictable interview regressions.
Wrong move: Loop endpoints miss first/last candidate.
Usually fails on: Fails on minimal arrays and exact-boundary answers.
Fix: Re-derive loops from inclusive/exclusive ranges before coding.