Problem Description

You are given a string s and an array of queries. Each query is represented as queries[i] = [left_i, right_i, k_i] and asks: can the substring s[left_i...right_i] be made into a palindrome by performing the following operations?

1. Rearrange the characters in the substring in any order
2. Replace up to k_i characters with any lowercase English letter

For each query, you need to determine if it's possible to create a palindrome using these operations and return true if possible, false otherwise.

Key Points:

• Each character replacement is counted individually. For example, if the substring is "aaa" and k = 2, you can only replace 2 of the 'a' characters
• The original string s is never modified - each query is independent
• Characters can be rearranged before replacements are made

Understanding Palindrome Formation: To form a palindrome from a substring with replacements:

• Characters that appear an even number of times can be paired and placed symmetrically
• Characters that appear an odd number of times need special handling:
  • At most one character can have an odd count (it goes in the middle)
  • If there are multiple characters with odd counts, we need to replace some to make them even
  • The minimum replacements needed is ⌊odd_count/2⌋ where odd_count is the number of characters with odd frequency

Example Walkthrough: For s = "abcda" and query [0, 3, 2] (substring "abcd" with up to 2 replacements):

• Character frequencies: a=1, b=1, c=1, d=1 (all odd)
• We have 4 characters with odd counts
• Minimum replacements needed: ⌊4/2⌋ = 2
• Since 2 ≤ 2, we can form a palindrome (e.g., change to "abba")

Intuition

The key insight is that to form a palindrome, characters must be paired symmetrically, except for at most one character that can sit in the middle (for odd-length palindromes).

When we count character frequencies in a substring:

• Characters with even counts are already "palindrome-ready" - they can be split into pairs and placed symmetrically
• Characters with odd counts are problematic - only one can remain odd (in the middle), the rest must be made even

For example, if we have the substring "aabbbcc":

• a appears 2 times (even) ✓
• b appears 3 times (odd) ✗
• c appears 2 times (even) ✓

We have 1 character with odd count. This can form a palindrome like "cabbbac" without any replacements.

But what if we have "abcd" where all 4 characters appear once (odd)? We need to convert some odd counts to even counts through replacements. The optimal strategy is to replace characters in pairs - replacing one occurrence of a character reduces its odd count by 1, making it even.

If we have x characters with odd counts:

• We can keep 1 character odd (for the middle position)
• The remaining x - 1 characters need to become even
• Each replacement fixes one odd-count character
• So we need at least ⌊x/2⌋ replacements

This explains why the solution checks if cnt // 2 <= k, where cnt is the number of characters with odd counts.

To efficiently count character frequencies for any substring [l, r], we use prefix sums. Instead of recounting characters for each query, we precompute cumulative counts: ss[i][j] stores how many times character j appears in s[0...i-1]. Then the count of character j in substring [l, r] is simply ss[r+1][j] - ss[l][j].

Learn more about Prefix Sum patterns.

Solution Approach

The solution uses prefix sum to efficiently count character frequencies in any substring. Here's the step-by-step implementation:

1. Build Prefix Sum Array

We create a 2D array ss where ss[i][j] represents the count of character j in the first i characters of string s:

ss = [[0] * 26 for _ in range(n + 1)]
for i, c in enumerate(s, 1):
    ss[i] = ss[i - 1][:]  # Copy previous counts
    ss[i][ord(c) - ord("a")] += 1  # Increment current character

• Initialize ss[0] with all zeros (empty prefix)
• For each position i, copy the counts from ss[i-1]
• Increment the count for the current character c
• Convert character to index using ord(c) - ord("a") (maps 'a'→0, 'b'→1, ..., 'z'→25)

2. Process Each Query

For each query [l, r, k]:

cnt = sum((ss[r + 1][j] - ss[l][j]) & 1 for j in range(26))
ans.append(cnt // 2 <= k)

• Calculate character frequency: For character j, its count in substring s[l...r] is ss[r + 1][j] - ss[l][j]
• Count odd frequencies: Use & 1 (bitwise AND with 1) to check if a count is odd
  • Even numbers: even & 1 = 0
  • Odd numbers: odd & 1 = 1
• Sum odd counts: cnt stores the total number of characters with odd frequencies
• Check palindrome possibility: We need ⌊cnt/2⌋ replacements to make a palindrome
  • If cnt // 2 <= k, we can form a palindrome with at most k replacements

Time Complexity: O(n + q) where n is the length of string and q is the number of queries

• Building prefix sum: O(n × 26) = O(n)
• Processing each query: O(26) = O(1)
• Total for all queries: O(q)

Space Complexity: O(n) for the prefix sum array ss with dimensions (n+1) × 26

Example Walkthrough

Let's walk through the solution with s = "bdbc" and query [1, 3, 1] (checking substring "dbc" with up to 1 replacement).

Step 1: Build Prefix Sum Array

We create a prefix sum array to track character counts up to each position:

Position:  0    1    2    3    4
String:    ""   "b"  "bd" "bdb" "bdbc"

For each position, we track counts of all 26 letters. Here are the relevant ones:

• ss[0]: b=0, c=0, d=0 (empty prefix)
• ss[1]: b=1, c=0, d=0 (after "b")
• ss[2]: b=1, c=0, d=1 (after "bd")
• ss[3]: b=2, c=0, d=1 (after "bdb")
• ss[4]: b=2, c=1, d=1 (after "bdbc")

Step 2: Process Query [1, 3, 1]

We want substring s[1...3] = "dbc" with k=1 replacement allowed.

Calculate character frequencies in substring using prefix sums:

• b: ss[4][1] - ss[1][1] = 2 - 1 = 1 (odd)
• c: ss[4][2] - ss[1][2] = 1 - 0 = 1 (odd)
• d: ss[4][3] - ss[1][3] = 1 - 0 = 1 (odd)

Count characters with odd frequencies:

• b appears 1 time (odd) → contributes 1
• c appears 1 time (odd) → contributes 1
• d appears 1 time (odd) → contributes 1
• Total: cnt = 3

Step 3: Check Palindrome Possibility

With 3 characters having odd counts, we need ⌊3/2⌋ = 1 replacement to form a palindrome.

Since 1 ≤ 1 (needed replacements ≤ k), the answer is true.

We could form palindrome "cbc" by replacing the 'd' with 'c', then rearranging.

Solution Implementation

Time and Space Complexity

Time Complexity: O((n + m) × 26) or O((n + m) × C) where C = 26

The time complexity breaks down into two main parts:

1. Building the prefix sum array: The outer loop runs n times (for each character in string s). For each iteration, we copy the previous row which takes O(26) time and update one position. This gives us O(n × 26) time.
2. Processing queries: For each of the m queries, we calculate the count of characters with odd frequencies by iterating through all 26 letters. This takes O(m × 26) time.

Therefore, the total time complexity is O(n × 26 + m × 26) = O((n + m) × 26).

Space Complexity: O(n × 26) or O(n × C) where C = 26

The space complexity is dominated by the prefix sum array ss, which has dimensions (n + 1) × 26. This stores the cumulative count of each of the 26 lowercase letters at each position from 0 to n. The answer list ans takes O(m) space, but since m could be smaller than n × 26, the overall space complexity remains O(n × 26).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Off-by-One Errors in Prefix Sum Indexing

A frequent mistake is incorrectly calculating the substring frequency using prefix sums. Developers often write:

# WRONG: This calculates s[left-1:right] instead of s[left:right+1]
char_frequency = prefix_counts[right][char_index] - prefix_counts[left][char_index]

Why it happens: The confusion arises because:

• String indexing is 0-based: s[left] is the character at position left
• Prefix sum prefix_counts[i] contains counts for s[0:i] (first i characters)
• To get substring s[left:right+1], we need prefix_counts[right+1] - prefix_counts[left]

Solution: Always remember that for substring s[left...right] (inclusive):

# CORRECT: Gets frequency in s[left:right+1]
char_frequency = prefix_counts[right + 1][char_index] - prefix_counts[left][char_index]

2. Misunderstanding the Replacement Count

Another common error is incorrectly calculating how many replacements are needed:

# WRONG: Thinks we need to replace all odd-count characters
replacements_needed = odd_count

Why it happens: The intuition that "all odd-count characters need fixing" misses the key insight that:

• One character with odd count can stay in the middle of the palindrome
• We only need to pair up the excess odd-count characters
• Each replacement fixes two odd-count characters (changes one odd to even, allowing pairing)

Solution: The correct formula is:

# CORRECT: Only need to fix pairs of odd-count characters
replacements_needed = odd_count // 2

3. Inefficient Character Frequency Calculation

Without prefix sums, a naive approach recalculates frequencies for each query:

# INEFFICIENT: O(q * n) time complexity
for left, right, k in queries:
    freq = [0] * 26
    for i in range(left, right + 1):
        freq[ord(s[i]) - ord('a')] += 1
    # ... rest of logic

Why it happens: It seems straightforward to count characters for each query independently, but this becomes prohibitively slow for large inputs.

Solution: Use prefix sums for O(1) frequency calculation per query:

# EFFICIENT: O(n) preprocessing, then O(1) per query
# Build prefix sums once
prefix_counts = [[0] * 26 for _ in range(n + 1)]
for i, c in enumerate(s, 1):
    prefix_counts[i] = prefix_counts[i - 1][:]
    prefix_counts[i][ord(c) - ord('a')] += 1

# Use prefix sums for each query
for left, right, k in queries:
    # O(26) = O(1) to calculate all frequencies
    for j in range(26):
        freq = prefix_counts[right + 1][j] - prefix_counts[left][j]

4. Forgetting to Handle Edge Cases

Some implementations fail on edge cases like:

• Single character substrings (always palindromes)
• Empty queries list
• k = 0 (no replacements allowed)

Solution: The provided algorithm naturally handles these cases:

• Single character: odd_count = 1, needs 1//2 = 0 replacements ✓
• No replacements (k=0): Only palindromes with at most 1 odd-count character work ✓
• Empty queries: Returns empty list ✓