Valid Palindrome III

Valid Palindrome III - Problem

Given a string s and an integer k, return true if s is a k-palindrome.

A string is k-palindrome if it can be transformed into a palindrome by removing at most k characters from it.

Example:

For string "abcdeca" and k = 2, we can remove characters at positions 1 and 4 ('b' and 'e') to get "acdca", which can be rearranged to form palindrome "aceca". However, this problem asks if we can form a palindrome by removing characters while maintaining their relative order.

Input & Output

Example 1 — Basic k-palindrome

$ Input: s = "abcdeca", k = 2

› Output: true

💡 Note: We can remove 'b' and 'd' to get "aceca" which is a palindrome. The minimum deletions needed is 2, which equals k=2, so return true.

Example 2 — Already palindrome

$ Input: s = "abcba", k = 1

› Output: true

💡 Note: String is already a palindrome, needs 0 deletions which is ≤ k=1.

Example 3 — Not enough deletions

$ Input: s = "abc", k = 1

› Output: false

💡 Note: To make "abc" a palindrome, we need to delete at least 2 characters (e.g., delete 'b' and 'c' to get "a"). Since 2 > k=1, return false.

Constraints

1 ≤ s.length ≤ 1000
0 ≤ k ≤ s.length
s consists only of lowercase English letters

Visualization

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Asked in

G Google 35 a Amazon 28 M Microsoft 22

The key insight is to find the minimum number of character deletions needed to make the string a palindrome, then check if it's ≤ k. Best approach uses 2D Dynamic Programming with bottom-up table building. Time: O(n²), Space: O(n²)

Common Approaches

✓ 2D Dynamic Programming

⏱️ Time: O(n²) Space: O(n²)

Use 2D DP table where dp[i][j] represents minimum deletions needed to make substring s[i...j] a palindrome. Build the solution bottom-up from smaller substrings to larger ones.

Brute Force - Try All Combinations

⏱️ Time: O(2^n × n) Space: O(n)

Generate every possible subsequence of the string by trying all combinations of character removals. For each subsequence, check if it's a palindrome and if the number of removed characters is at most k.

Memoization - Top-Down DP

⏱️ Time: O(n²) Space: O(n²)

Use top-down dynamic programming to find the minimum number of characters that need to be deleted to make the string a palindrome. Cache results to avoid recomputation.

2D Dynamic Programming — Algorithm Steps

Step 1: Create DP table dp[i][j] for all substring ranges
Step 2: Base case: single characters and empty strings need 0 deletions
Step 3: For each substring, if ends match, use inner result; else try both deletions
Step 4: Return dp[0][n-1] ≤ k

Visualization

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Step-by-Step Walkthrough

Initialize Table

Create n×n table, single chars need 0 deletions

Fill by Length

Process substrings by increasing length

Compare & Decide

If chars match use inner result, else try both deletions

Code -

solution.c — C

#include <stdio.h>
#include <string.h>
#include <stdbool.h>
#include <stdlib.h>

bool solution(char* s, int k) {
    int n = strlen(s);
    if (n == 0) return true;
    
    // dp[i][j] = min deletions to make s[i...j] palindrome
    int dp[1000][1000];
    memset(dp, 0, sizeof(dp));
    
    // Fill DP table for all substring lengths
    for (int length = 2; length <= n; length++) { // length 1 already 0
        for (int i = 0; i <= n - length; i++) {
            int j = i + length - 1;
            if (s[i] == s[j]) {
                dp[i][j] = dp[i + 1][j - 1];
            } else {
                // Try deleting left or right character
                int deleteLeft = dp[i + 1][j];
                int deleteRight = dp[i][j - 1];
                dp[i][j] = 1 + (deleteLeft < deleteRight ? deleteLeft : deleteRight);
            }
        }
    }
    
    return dp[0][n - 1] <= k;
}

int main() {
    char s[1001];
    int k;
    
    fgets(s, sizeof(s), stdin);
    s[strcspn(s, "\n")] = 0;
    scanf("%d", &k);
    
    bool result = solution(s, k);
    printf(result ? "true\n" : "false\n");
    
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n²)

Nested loops fill n×n DP table, each cell computed in O(1)

⚠ Quadratic Growth

Space Complexity

O(n²)

2D DP table stores results for all possible substring ranges

⚠ Quadratic Space

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Constraints

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Related Problems

Common Approaches

2D Dynamic Programming — Algorithm Steps

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Code -

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