Minimum insertions to form a palindrome
Last Updated : 09 May, 2025
Given a string s, the task is to find the minimum number of characters to be inserted to convert it to a palindrome.
Examples:
Input: s = “geeks”
Output: 3
Explanation: “skgeegks” is a palindromic string, which requires 3 insertions.
Input: s= “abcd”
Output: 3
Explanation: “abcdcba” is a palindromic string.
Input: s= “aba”
Output: 0
Explanation: Given string is already a palindrome hence no insertions are required.
Using Recursion – O(2^n) Time and O(n) Space
The minimum number of insertions in the string str[l…..h] can be given as:
- if str[l] is equal to str[h] findMinInsertions(str[l+1…..h-1])
- otherwise, min(findMinInsertions(str[l…..h-1]), findMinInsertions(str[l+1…..h])) + 1
C++ // C++ approach to find Minimum // insertions to form a palindrome #include<bits/stdc++.h> using namespace std; // Recursive function to find // minimum number of insertions int minRecur(string &s, int l, int h) { // Base case if (l >= h) return 0; // if first and last char are same // then no need to insert element if(s[l] == s[h]) { return minRecur(s, l + 1, h - 1); } // Insert at begining or insert at end return min(minRecur(s, l + 1, h), minRecur(s, l, h - 1)) + 1; } int findMinInsertions(string &s) { return minRecur(s, 0, s.size() - 1); } int main() { string s = "geeks"; cout << findMinInsertions(s); return 0; } // Java approach to find Minimum // insertions to form a palindrome class GfG { // Recursive function to find // minimum number of insertions static int minRecur(String s, int l, int h) { // Base case if (l >= h) return 0; // if first and last char are same // then no need to insert element if (s.charAt(l) == s.charAt(h)) { return minRecur(s, l + 1, h - 1); } // Insert at begining or insert at end return Math.min(minRecur(s, l + 1, h), minRecur(s, l, h - 1)) + 1; } static int findMinInsertions(String s) { return minRecur(s, 0, s.length() - 1); } public static void main(String[] args) { String s = "geeks"; System.out.println(findMinInsertions(s)); } } # Python approach to find Minimum # insertions to form a palindrome # Recursive function to find # minimum number of insertions def minRecur(s, l, h): # Base case if l >= h: return 0 # if first and last char are same # then no need to insert element if s[l] == s[h]: return minRecur(s, l + 1, h - 1) # Insert at begining or insert at end return min(minRecur(s, l + 1, h), minRecur(s, l, h - 1)) + 1 def findMinInsertions(s): return minRecur(s, 0, len(s) - 1) if __name__ == "__main__": s = "geeks" print(findMinInsertions(s))
// C# approach to find Minimum // insertions to form a palindrome using System; class GfG { // Recursive function to find // minimum number of insertions static int minRecur(string s, int l, int h) { // Base case if (l >= h) return 0; // if first and last char are same // then no need to insert element if (s[l] == s[h]) { return minRecur(s, l + 1, h - 1); } // Insert at begining or insert at end return Math.Min(minRecur(s, l + 1, h), minRecur(s, l, h - 1)) + 1; } static int findMinInsertions(string s) { return minRecur(s, 0, s.Length - 1); } static void Main(string[] args) { string s = "geeks"; Console.WriteLine(findMinInsertions(s)); } } // JavaScript approach to find Minimum // insertions to form a palindrome // Recursive function to find // minimum number of insertions function minRecur(s, l, h) { // Base case if (l >= h) return 0; // if first and last char are same // then no need to insert element if (s[l] === s[h]) { return minRecur(s, l + 1, h - 1); } // Insert at begining or insert at end return Math.min(minRecur(s, l + 1, h), minRecur(s, l, h - 1)) + 1; } function findMinInsertions(s) { return minRecur(s, 0, s.length - 1); } const s = "geeks"; console.log(findMinInsertions(s)); Using Top-Down DP (Memoization) – O(n^2) Time and O(n^2) Space
If we notice carefully, we can observe that the above recursive solution holds the following two properties of Dynamic Programming:
1. Optimal Substructure: The minimum number of insertions required to make the substring str[l…h] a palindrome depends on the optimal solutions of its subproblems. If str[l] is equal to str[h] then we call recursive call on l + 1 and h – 1 other we make two recursive call to get optimal answer.
2. Overlapping Subproblems: When using a recursive approach, we notice that certain subproblems are computed multiple times. To avoid this redundancy, we can use memoization by storing the results of already computed subproblems in a 2D array.
The problem involves changing two parameters: l (starting index) and h (ending index). Hence, we need to create a 2D array of size (n x n) to store the results, where n is the length of the string. By using this 2D array, we can efficiently reuse previously computed results and optimize the solution.
C++ // C++ approach to find Minimum // insertions to form a palindrome #include<bits/stdc++.h> using namespace std; // Recursive function to find // minimum number of insertions int minRecur(string &s, int l, int h, vector<vector<int>> &memo) { // Base case if (l >= h) return 0; // If value is memoized if (memo[l][h] != -1) return memo[l][h]; // if first and last char are same // then no need to insert element if(s[l] == s[h]) { return memo[l][h] = minRecur(s, l + 1, h - 1, memo); } // Insert at begining or insert at end return memo[l][h] = min(minRecur(s, l + 1, h, memo), minRecur(s, l, h - 1, memo)) + 1; } int findMinInsertions(string &s) { int n = s.length(); vector<vector<int>> memo(n, vector<int>(n, -1)); return minRecur(s, 0, n - 1, memo); } int main() { string s = "geeks"; cout << findMinInsertions(s); return 0; } // Java approach to find Minimum // insertions to form a palindrome class GfG { // Recursive function to find // minimum number of insertions static int minRecur(String s, int l, int h, int[][] memo) { // Base case if (l >= h) return 0; // If value is memoized if (memo[l][h] != -1) return memo[l][h]; // if first and last char are same // then no need to insert element if (s.charAt(l) == s.charAt(h)) { return memo[l][h] = minRecur(s, l + 1, h - 1, memo); } // Insert at begining or insert at end return memo[l][h] = Math.min(minRecur(s, l + 1, h, memo), minRecur(s, l, h - 1, memo)) + 1; } static int findMinInsertions(String s) { int n = s.length(); int[][] memo = new int[n][n]; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { memo[i][j] = -1; } } return minRecur(s, 0, n - 1, memo); } public static void main(String[] args) { String s = "geeks"; System.out.println(findMinInsertions(s)); } } # Python approach to find Minimum # insertions to form a palindrome # Recursive function to find # minimum number of insertions def minRecur(s, l, h, memo): # Base case if l >= h: return 0 # If value is memoized if memo[l][h] != -1: return memo[l][h] # if first and last char are same # then no need to insert element if s[l] == s[h]: memo[l][h] = minRecur(s, l + 1, h - 1, memo) return memo[l][h] # Insert at begining or insert at end memo[l][h] = min(minRecur(s, l + 1, h, memo), minRecur(s, l, h - 1, memo)) + 1 return memo[l][h] def findMinInsertions(s): n = len(s) memo = [[-1] * n for _ in range(n)] return minRecur(s, 0, n - 1, memo) if __name__ == "__main__": s = "geeks" print(findMinInsertions(s))
// C# approach to find Minimum // insertions to form a palindrome using System; class GfG { // Recursive function to find // minimum number of insertions static int minRecur(string s, int l, int h, int[,] memo) { // Base case if (l >= h) return 0; // If value is memoized if (memo[l, h] != -1) return memo[l, h]; // if first and last char are same // then no need to insert element if (s[l] == s[h]) { return memo[l, h] = minRecur(s, l + 1, h - 1, memo); } // Insert at begining or insert at end return memo[l, h] = Math.Min(minRecur(s, l + 1, h, memo), minRecur(s, l, h - 1, memo)) + 1; } static int findMinInsertions(string s) { int n = s.Length; int[,] memo = new int[n, n]; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { memo[i, j] = -1; } } return minRecur(s, 0, n - 1, memo); } static void Main(string[] args) { string s = "geeks"; Console.WriteLine(findMinInsertions(s)); } } // JavaScript approach to find Minimum // insertions to form a palindrome // Recursive function to find // minimum number of insertions function minRecur(s, l, h, memo) { // Base case if (l >= h) return 0; // If value is memoized if (memo[l][h] !== -1) return memo[l][h]; // if first and last char are same // then no need to insert element if (s[l] === s[h]) { return memo[l][h] = minRecur(s, l + 1, h - 1, memo); } // Insert at begining or insert at end return memo[l][h] = Math.min(minRecur(s, l + 1, h, memo), minRecur(s, l, h - 1, memo)) + 1; } function findMinInsertions(s) { const n = s.length; const memo = Array.from({ length: n }, () => Array(n).fill(-1)); return minRecur(s, 0, n - 1, memo); } const s = "geeks"; console.log(findMinInsertions(s)); Using Bottom-Up DP (Tabulation) – O(n^2) Time and O(n^2) Space
The iterative approach for finding the minimum number of insertions needed to make a string palindrome follows a similar concept to the recursive solution but builds the solution in a bottom-up manner using a 2D dynamic programming table.
Base Case:
- if l == h than dp[l][h] = 0. Where 0 <= l <= n – 1
- If the substring has two characters (h == l + 1), we check if they are the same. If they are, dp[l][h] = 0, otherwise, dp[l][h] = 1.
Recursive Case:
- If str[l] == str[h], then dp[l][h] = dp[l+1][h-1].
- Otherwise, dp[l][h] = min(dp[l+1][h], dp[l][h-1]) + 1.
C++ // C++ approach to find Minimum // insertions to form a palindrome #include<bits/stdc++.h> using namespace std; // Function to find minimum insertions // to make string palindrome int findMinInsertions(string &s) { int n = s.size(); // dp[i][j] will store the minimum number of insertions needed // to convert str[i..j] into a palindrome vector<vector<int>> dp(n, vector<int>(n, 0)); // len is the length of the substring for (int len = 2; len <= n; len++) { for (int l = 0; l <= n - len; l++) { // ending index of the current substring int h = l + len - 1; // If the characters at both ends are // equal, no new insertions needed if (s[l] == s[h]) { dp[l][h] = dp[l + 1][h - 1]; } else { // Insert at the beginning or at the end dp[l][h] = min(dp[l + 1][h], dp[l][h - 1]) + 1; } } } // The result is in dp[0][n-1] which // represents the entire string return dp[0][n - 1]; } int main() { string s = "geeks"; cout << findMinInsertions(s); return 0; } // Java approach to find Minimum // insertions to form a palindrome class GfG { // Function to find minimum insertions // to make string palindrome static int findMinInsertions(String s) { int n = s.length(); // dp[i][j] will store the minimum number of insertions needed // to convert str[i..j] into a palindrome int[][] dp = new int[n][n]; // len is the length of the substring for (int len = 2; len <= n; len++) { for (int l = 0; l <= n - len; l++) { // ending index of the current substring int h = l + len - 1; // If the characters at both ends are // equal, no new insertions needed if (s.charAt(l) == s.charAt(h)) { dp[l][h] = dp[l + 1][h - 1]; } else { // Insert at the beginning or at the end dp[l][h] = Math.min(dp[l + 1][h], dp[l][h - 1]) + 1; } } } // The result is in dp[0][n-1] which // represents the entire string return dp[0][n - 1]; } public static void main(String[] args) { String s = "geeks"; System.out.println(findMinInsertions(s)); } } # Python approach to find Minimum # insertions to form a palindrome # Function to find minimum insertions # to make string palindrome def findMinInsertions(s): n = len(s) # dp[i][j] will store the minimum number of insertions needed # to convert str[i..j] into a palindrome dp = [[0] * n for _ in range(n)] # len is the length of the substring for length in range(2, n + 1): for l in range(n - length + 1): # ending index of the current substring h = l + length - 1 # If the characters at both ends are # equal, no new insertions needed if s[l] == s[h]: dp[l][h] = dp[l + 1][h - 1] else: # Insert at the beginning or at the end dp[l][h] = min(dp[l + 1][h], dp[l][h - 1]) + 1 # The result is in dp[0][n-1] which # represents the entire string return dp[0][n - 1] if __name__ == "__main__": s = "geeks" print(findMinInsertions(s))
// C# approach to find Minimum // insertions to form a palindrome using System; class MainClass { // Function to find minimum insertions // to make string palindrome static int findMinInsertions(string s) { int n = s.Length; // dp[i,j] will store the minimum number of insertions needed // to convert str[i..j] into a palindrome int[,] dp = new int[n, n]; // len is the length of the substring for (int len = 2; len <= n; len++) { for (int l = 0; l <= n - len; l++) { // ending index of the current substring int h = l + len - 1; // If the characters at both ends are // equal, no new insertions needed if (s[l] == s[h]) { dp[l, h] = dp[l + 1, h - 1]; } else { // Insert at the beginning or at the end dp[l, h] = Math.Min(dp[l + 1, h], dp[l, h - 1]) + 1; } } } // The result is in dp[0,n-1] which // represents the entire string return dp[0, n - 1]; } static void Main(string[] args) { string s = "geeks"; Console.WriteLine(findMinInsertions(s)); } } // JavaScript approach to find Minimum // insertions to form a palindrome // Function to find minimum insertions // to make string palindrome function findMinInsertions(s) { const n = s.length; // dp[i][j] will store the minimum number of insertions needed // to convert str[i..j] into a palindrome const dp = Array.from({ length: n }, () => Array(n).fill(0)); // len is the length of the substring for (let len = 2; len <= n; len++) { for (let l = 0; l <= n - len; l++) { // ending index of the current substring const h = l + len - 1; // If the characters at both ends are // equal, no new insertions needed if (s[l] === s[h]) { dp[l][h] = dp[l + 1][h - 1]; } else { // Insert at the beginning or at the end dp[l][h] = Math.min(dp[l + 1][h], dp[l][h - 1]) + 1; } } } // The result is in dp[0][n-1] which // represents the entire string return dp[0][n - 1]; } const s = "geeks"; console.log(findMinInsertions(s)); Using Space Optimized DP – O(n ^ 2) Time and O(n) Space
The idea is to reuse the array in such a way that we store the results for the previous row in the same array while iterating through the columns.
C++ // C++ approach to find Minimum // insertions to form a palindrome #include<bits/stdc++.h> using namespace std; // Function to find minimum insertions // to make string palindrome int findMinInsertions(string &s) { int n = s.size(); vector<int> dp(n, 0); // Iterate over each character from right to left for (int l = n - 2; l >= 0; l--) { // This represents dp[l+1][h-1] from the previous row int prev = 0; for (int h = l + 1; h < n; h++) { // Save current dp[h] before overwriting int temp = dp[h]; if (s[l] == s[h]) { // No new insertion needed if characters match dp[h] = prev; } else { // Take min of two cases + 1 dp[h] = min(dp[h], dp[h - 1]) + 1; } // Update prev for the next column prev = temp; } } return dp[n - 1]; } int main() { string s = "geeks"; cout << findMinInsertions(s); return 0; } // Java approach to find Minimum // insertions to form a palindrome class GfG { // Function to find minimum insertions // to make string palindrome static int findMinInsertions(String s) { int n = s.length(); int[] dp = new int[n]; // Iterate over each character from right to left for (int l = n - 2; l >= 0; l--) { // This represents dp[l+1][h-1] from the previous row int prev = 0; for (int h = l + 1; h < n; h++) { // Save current dp[h] before overwriting int temp = dp[h]; if (s.charAt(l) == s.charAt(h)) { // No new insertion needed if characters match dp[h] = prev; } else { // Take min of two cases + 1 dp[h] = Math.min(dp[h], dp[h - 1]) + 1; } // Update prev for the next column prev = temp; } } return dp[n - 1]; } public static void main(String[] args) { String s = "geeks"; System.out.println(findMinInsertions(s)); } } # Python approach to find Minimum # insertions to form a palindrome # Function to find minimum insertions # to make string palindrome def findMinInsertions(s): n = len(s) dp = [0] * n # Iterate over each character from right to left for l in range(n - 2, -1, -1): # This represents dp[l+1][h-1] from the previous row prev = 0 for h in range(l + 1, n): # Save current dp[h] before overwriting temp = dp[h] if s[l] == s[h]: # No new insertion needed if characters match dp[h] = prev else: # Take min of two cases + 1 dp[h] = min(dp[h], dp[h - 1]) + 1 # Update prev for the next column prev = temp return dp[n - 1] if __name__ == "__main__": s = "geeks" print(findMinInsertions(s))
// C# approach to find Minimum // insertions to form a palindrome using System; class GfG { // Function to find minimum insertions // to make string palindrome static int findMinInsertions(string s) { int n = s.Length; int[] dp = new int[n]; // Iterate over each character from right to left for (int l = n - 2; l >= 0; l--) { // This represents dp[l+1][h-1] from the previous row int prev = 0; for (int h = l + 1; h < n; h++) { // Save current dp[h] before overwriting int temp = dp[h]; if (s[l] == s[h]) { // No new insertion needed if characters match dp[h] = prev; } else { // Take min of two cases + 1 dp[h] = Math.Min(dp[h], dp[h - 1]) + 1; } // Update prev for the next column prev = temp; } } return dp[n - 1]; } static void Main(string[] args) { string s = "geeks"; Console.WriteLine(findMinInsertions(s)); } } // JavaScript approach to find Minimum // insertions to form a palindrome // Function to find minimum insertions // to make string palindrome function findMinInsertions(s) { const n = s.length; const dp = new Array(n).fill(0); // Iterate over each character from right to left for (let l = n - 2; l >= 0; l--) { // This represents dp[l+1][h-1] from the previous row let prev = 0; for (let h = l + 1; h < n; h++) { // Save current dp[h] before overwriting let temp = dp[h]; if (s[l] === s[h]) { // No new insertion needed if characters match dp[h] = prev; } else { // Take min of two cases + 1 dp[h] = Math.min(dp[h], dp[h - 1]) + 1; } // Update prev for the next column prev = temp; } } return dp[n - 1]; } const s = "geeks"; console.log(findMinInsertions(s)); Related article:
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