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));
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