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 bit manipulation strategy.
You are given two 0-indexed binary strings s and target of the same length n. You can do the following operation on s any number of times:
i and j where 0 <= i, j < n.s[i] with (s[i] OR s[j]) and s[j] with (s[i] XOR s[j]).For example, if s = "0110", you can choose i = 0 and j = 2, then simultaneously replace s[0] with (s[0] OR s[2] = 0 OR 1 = 1), and s[2] with (s[0] XOR s[2] = 0 XOR 1 = 1), so we will have s = "1110".
Return true if you can make the string s equal to target, or false otherwise.
Example 1:
Input: s = "1010", target = "0110" Output: true Explanation: We can do the following operations: - Choose i = 2 and j = 0. We have now s = "0010". - Choose i = 2 and j = 1. We have now s = "0110". Since we can make s equal to target, we return true.
Example 2:
Input: s = "11", target = "00" Output: false Explanation: It is not possible to make s equal to target with any number of operations.
Constraints:
n == s.length == target.length2 <= n <= 105s and target consist of only the digits 0 and 1.Problem summary: You are given two 0-indexed binary strings s and target of the same length n. You can do the following operation on s any number of times: Choose two different indices i and j where 0 <= i, j < n. Simultaneously, replace s[i] with (s[i] OR s[j]) and s[j] with (s[i] XOR s[j]). For example, if s = "0110", you can choose i = 0 and j = 2, then simultaneously replace s[0] with (s[0] OR s[2] = 0 OR 1 = 1), and s[2] with (s[0] XOR s[2] = 0 XOR 1 = 1), so we will have s = "1110". Return true if you can make the string s equal to target, or false otherwise.
Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.
Pattern signal: Bit Manipulation
"1010" "0110"
"11" "00"
minimum-one-bit-operations-to-make-integers-zero)Source-backed implementations are provided below for direct study and interview prep.
// Accepted solution for LeetCode #2546: Apply Bitwise Operations to Make Strings Equal
class Solution {
public boolean makeStringsEqual(String s, String target) {
return s.contains("1") == target.contains("1");
}
}
// Accepted solution for LeetCode #2546: Apply Bitwise Operations to Make Strings Equal
func makeStringsEqual(s string, target string) bool {
return strings.Contains(s, "1") == strings.Contains(target, "1")
}
# Accepted solution for LeetCode #2546: Apply Bitwise Operations to Make Strings Equal
class Solution:
def makeStringsEqual(self, s: str, target: str) -> bool:
return ("1" in s) == ("1" in target)
// Accepted solution for LeetCode #2546: Apply Bitwise Operations to Make Strings Equal
impl Solution {
pub fn make_strings_equal(s: String, target: String) -> bool {
s.contains('1') == target.contains('1')
}
}
// Accepted solution for LeetCode #2546: Apply Bitwise Operations to Make Strings Equal
function makeStringsEqual(s: string, target: string): boolean {
return s.includes('1') === target.includes('1');
}
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.