Minimum insertions to form a palindrome
Last Updated :
23 Jul, 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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Bitwise AlgorithmsBitwise algorithms in Data Structures and Algorithms (DSA) involve manipulating individual bits of binary representations of numbers to perform operations efficiently. These algorithms utilize bitwise operators like AND, OR, XOR, NOT, Left Shift, and Right Shift.BasicsIntroduction to Bitwise Algorit
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Advanced
Segment TreeSegment Tree is a data structure that allows efficient querying and updating of intervals or segments of an array. It is particularly useful for problems involving range queries, such as finding the sum, minimum, maximum, or any other operation over a specific range of elements in an array. The tree
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Pattern SearchingPattern searching algorithms are essential tools in computer science and data processing. These algorithms are designed to efficiently find a particular pattern within a larger set of data. Patten SearchingImportant Pattern Searching Algorithms:Naive String Matching : A Simple Algorithm that works i
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GeometryGeometry is a branch of mathematics that studies the properties, measurements, and relationships of points, lines, angles, surfaces, and solids. From basic lines and angles to complex structures, it helps us understand the world around us.Geometry for Students and BeginnersThis section covers key br
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