LeetCode #1177 — MEDIUM

Can Make Palindrome from Substring

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

Problem Statement

You are given a string s and array queries where queries[i] = [left_i, right_i, k_i]. We may rearrange the substring s[left_i...right_i] for each query and then choose up to k_i of them to replace with any lowercase English letter.

If the substring is possible to be a palindrome string after the operations above, the result of the query is true. Otherwise, the result is false.

Return a boolean array answer where answer[i] is the result of the i^th query queries[i].

Note that each letter is counted individually for replacement, so if, for example s[left_i...right_i] = "aaa", and k_i = 2, we can only replace two of the letters. Also, note that no query modifies the initial string s.

Example :

Input: s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]
Output: [true,false,false,true,true]
Explanation:
queries[0]: substring = "d", is palidrome.
queries[1]: substring = "bc", is not palidrome.
queries[2]: substring = "abcd", is not palidrome after replacing only 1 character.
queries[3]: substring = "abcd", could be changed to "abba" which is palidrome. Also this can be changed to "baab" first rearrange it "bacd" then replace "cd" with "ab".
queries[4]: substring = "abcda", could be changed to "abcba" which is palidrome.

Example 2:

Input: s = "lyb", queries = [[0,1,0],[2,2,1]]
Output: [false,true]

Constraints:

1 <= s.length, queries.length <= 10⁵
0 <= left_i <= right_i < s.length
0 <= k_i <= s.length
s consists of lowercase English letters.

Step 01

Brute Force Baseline

Problem summary: You are given a string s and array queries where queries[i] = [lefti, righti, ki]. We may rearrange the substring s[lefti...righti] for each query and then choose up to ki of them to replace with any lowercase English letter. If the substring is possible to be a palindrome string after the operations above, the result of the query is true. Otherwise, the result is false. Return a boolean array answer where answer[i] is the result of the ith query queries[i]. Note that each letter is counted individually for replacement, so if, for example s[lefti...righti] = "aaa", and ki = 2, we can only replace two of the letters. Also, note that no query modifies the initial string s. Example : Input: s = "abcda", queries = [[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]] Output: [true,false,false,true,true] Explanation: queries[0]: substring = "d", is palidrome. queries[1]: substring = "bc", is not

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Bit Manipulation

Example 1

"abcda"
[[3,3,0],[1,2,0],[0,3,1],[0,3,2],[0,4,1]]

Example 2

"lyb"
[[0,1,0],[2,2,1]]

Core Insight

What unlocks the optimal approach

Since we can rearrange the substring, all we care about is the frequency of each character in that substring.
How to find the character frequencies efficiently ?
As a preprocess, calculate the accumulate frequency of all characters for all prefixes of the string.
How to check if a substring can be changed to a palindrome given its characters frequency ?
Count the number of odd frequencies, there can be at most one odd frequency in a palindrome.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1177: Can Make Palindrome from Substring
class Solution {
    public List<Boolean> canMakePaliQueries(String s, int[][] queries) {
        int n = s.length();
        int[][] ss = new int[n + 1][26];
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j < 26; ++j) {
                ss[i][j] = ss[i - 1][j];
            }
            ss[i][s.charAt(i - 1) - 'a']++;
        }
        List<Boolean> ans = new ArrayList<>();
        for (var q : queries) {
            int l = q[0], r = q[1], k = q[2];
            int x = 0;
            for (int j = 0; j < 26; ++j) {
                x += (ss[r + 1][j] - ss[l][j]) & 1;
            }
            ans.add(x / 2 <= k);
        }
        return ans;
    }
}

// Accepted solution for LeetCode #1177: Can Make Palindrome from Substring
func canMakePaliQueries(s string, queries [][]int) (ans []bool) {
	n := len(s)
	ss := make([][26]int, n+1)
	for i := 1; i <= n; i++ {
		for j := 0; j < 26; j++ {
			ss[i][j] = ss[i-1][j]
		}
		ss[i][s[i-1]-'a']++
	}
	for _, q := range queries {
		l, r, k := q[0], q[1], q[2]
		x := 0
		for j := 0; j < 26; j++ {
			x += (ss[r+1][j] - ss[l][j]) & 1
		}
		ans = append(ans, x/2 <= k)
	}
	return
}

# Accepted solution for LeetCode #1177: Can Make Palindrome from Substring
class Solution:
    def canMakePaliQueries(self, s: str, queries: List[List[int]]) -> List[bool]:
        n = len(s)
        ss = [[0] * 26 for _ in range(n + 1)]
        for i, c in enumerate(s, 1):
            ss[i] = ss[i - 1][:]
            ss[i][ord(c) - ord("a")] += 1
        ans = []
        for l, r, k in queries:
            cnt = sum((ss[r + 1][j] - ss[l][j]) & 1 for j in range(26))
            ans.append(cnt // 2 <= k)
        return ans

// Accepted solution for LeetCode #1177: Can Make Palindrome from Substring
struct Solution;

impl Solution {
    fn can_make_pali_queries(s: String, queries: Vec<Vec<i32>>) -> Vec<bool> {
        let n = s.len();
        let mut prefix: Vec<u32> = vec![0; n + 1];
        for (i, c) in s.char_indices() {
            prefix[i + 1] = prefix[i] ^ (1 << (c as u32 - 'a' as u32));
        }
        let mut res = vec![];
        for q in queries {
            let left = q[0] as usize;
            let right = q[1] as usize + 1;
            let k = q[2] as u32;
            res.push(k * 2 >= (prefix[right] ^ prefix[left]).count_ones() - 1);
        }
        res
    }
}

#[test]
fn test() {
    let s = "abcda".to_string();
    let queries = vec_vec_i32![[3, 3, 0], [1, 2, 0], [0, 3, 1], [0, 3, 2], [0, 4, 1]];
    let res = vec![true, false, false, true, true];
    assert_eq!(Solution::can_make_pali_queries(s, queries), res);
}

// Accepted solution for LeetCode #1177: Can Make Palindrome from Substring
function canMakePaliQueries(s: string, queries: number[][]): boolean[] {
    const n = s.length;
    const ss: number[][] = Array(n + 1)
        .fill(0)
        .map(() => Array(26).fill(0));
    for (let i = 1; i <= n; ++i) {
        ss[i] = ss[i - 1].slice();
        ++ss[i][s.charCodeAt(i - 1) - 97];
    }
    const ans: boolean[] = [];
    for (const [l, r, k] of queries) {
        let x = 0;
        for (let j = 0; j < 26; ++j) {
            x += (ss[r + 1][j] - ss[l][j]) & 1;
        }
        ans.push(x >> 1 <= k);
    }
    return ans;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O((n + m)

Space

O(n × C)

Approach Breakdown

SORT + SCAN

O(n log n) time

O(n) space

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.

BIT MANIPULATION

O(n) time

O(1) space

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.

Shortcut: Bit operations are O(1). XOR cancels duplicates. Single pass → O(n) time, O(1) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

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.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.