Palindrome by Front Insertion
Last Updated :
26 Jul, 2025
Given a string s consisting of only lowercase English letters, find the minimum number of characters that need to be added to the front of s to make it a palindrome.
Note: A palindrome is a string that reads the same forward and backward.
Examples:
Input: s = "abc"
Output: 2
Explanation: We can make above string palindrome as "cbabc", by adding 'b' and 'c' at front.
Input: s = "aacecaaaa"
Output: 2
Explanation: We can make above string palindrome as "aaaacecaaaa" by adding two a's at front of string.
[Naive Approach] Checking all prefixes - O(n^2) Time and O(1) Space
The idea is based on the observation that we need to find the longest prefix from given string which is also a palindrome. Then minimum front characters to be added to make given string palindrome will be the remaining characters.
C++
#include <iostream>
using namespace std;
// function to check if the substring s[i...j] is a palindrome
bool isPalindrome(string &s, int i, int j) {
while (i < j) {
// if characters at the ends are not equal,
// it's not a palindrome
if (s[i] != s[j]) {
return false;
}
i++;
j--;
}
return true;
}
int minChar(string &s) {
int cnt = 0;
int i = s.size() - 1;
// iterate from the end of the string, checking for the
// longestpalindrome starting from the beginning
while (i >= 0 && !isPalindrome(s, 0, i)) {
i--;
cnt++;
}
return cnt;
}
int main() {
string s = "aacecaaaa";
cout << minChar(s);
return 0;
}
#include <stdio.h>
#include <stdbool.h>
#include <string.h>
// function to check if the substring s[i...j] is a palindrome
bool isPalindrome(char s[], int i, int j) {
while (i < j) {
// if characters at the ends are not the same,
// it's not a palindrome
if (s[i] != s[j]) {
return false;
}
i++;
j--;
}
return true;
}
int minChar(char s[]) {
int cnt = 0;
int i = strlen(s) - 1;
// iterate from the end of the string, checking for the
// longest palindrome starting from the beginning
while (i >= 0 && !isPalindrome(s, 0, i)) {
i--;
cnt++;
}
return cnt;
}
int main() {
char s[] = "aacecaaaa";
printf("%d", minChar(s));
return 0;
}
class GfG {
// function to check if the substring
// s[i...j] is a palindrome
static boolean isPalindrome(String s, int i, int j) {
while (i < j) {
// if characters at the ends are not the same,
// it's not a palindrome
if (s.charAt(i) != s.charAt(j)) {
return false;
}
i++;
j--;
}
return true;
}
static int minChar(String s) {
int cnt = 0;
int i = s.length() - 1;
// iterate from the end of the string, checking for the
// longest palindrome starting from the beginning
while (i >= 0 && !isPalindrome(s, 0, i)) {
i--;
cnt++;
}
return cnt;
}
public static void main(String[] args) {
String s = "aacecaaaa";
System.out.println(minChar(s));
}
}
# function to check if the substring s[i...j] is a palindrome
def isPalindrome(s, i, j):
while i < j:
# if characters at the ends are not the same,
# it's not a palindrome
if s[i] != s[j]:
return False
i += 1
j -= 1
return True
def minChar(s):
cnt = 0
i = len(s) - 1
# iterate from the end of the string, checking for the
# longest palindrome starting from the beginning
while i >= 0 and not isPalindrome(s, 0, i):
i -= 1
cnt += 1
return cnt
if __name__ == "__main__":
s = "aacecaaaa"
print(minChar(s))
using System;
class GfG {
// function to check if the substring s[i...j] is a palindrome
static bool isPalindrome(string s, int i, int j) {
while (i < j) {
// if characters at the ends are not the same,
// it's not a palindrome
if (s[i] != s[j]) {
return false;
}
i++;
j--;
}
return true;
}
static int minChar(string s) {
int cnt = 0;
int i = s.Length - 1;
// iterate from the end of the string, checking for the longest
// palindrome starting from the beginning
while (i >= 0 && !isPalindrome(s, 0, i)) {
i--;
cnt++;
}
return cnt;
}
static void Main() {
string s = "aacecaaaa";
Console.WriteLine(minChar(s));
}
}
// function to check if the substring s[i...j] is a palindrome
function isPalindrome(s, i, j) {
while (i < j) {
// if characters at the ends are not the same,
// it's not a palindrome
if (s[i] !== s[j]) {
return false;
}
i++;
j--;
}
return true;
}
function minChar(s) {
let cnt = 0;
let i = s.length - 1;
// iterate from the end of the string, checking for the
// longest palindrome starting from the beginning
while (i >= 0 && !isPalindrome(s, 0, i)) {
i--;
cnt++;
}
return cnt;
}
// Driver code
let s = "aacecaaaa";
console.log(minChar(s));
[Expected Approach 1] Using lps array of KMP Algorithm - O(n) Time and O(n) Space
The key observation is that the longest palindromic prefix of a string becomes the longest palindromic suffix of its reverse.
Given a string s = "aacecaaaa", its reverse revS = "aaaacecaa". The longest palindromic prefix of s is "aacecaa".
To find this efficiently, we use the LPS array from the KMP algorithm. We concatenate the original string with a special character and its reverse: s + '$' + revS.
The LPS array for this combined string helps identify the longest prefix of s that matches a suffix of revS, which also represents the palindromic prefix of s.
The last value of the LPS array tells us how many characters already form a palindrome at the beginning. Thus, the minimum number of characters to add to make s a palindrome is s.length() - lps.back().
C++
#include <algorithm>
#include <iostream>
#include <vector>
using namespace std;
vector<int> computeLPSArray(string &pat) {
int n = pat.length();
vector<int> lps(n);
// lps[0] is always 0
lps[0] = 0;
int len = 0;
// loop calculates lps[i] for i = 1 to M-1
int i = 1;
while (i < n) {
// if the characters match, increment len
// and set lps[i]
if (pat[i] == pat[len]) {
len++;
lps[i] = len;
i++;
}
// if there is a mismatch
else {
// if len is not zero, update len to
// the last known prefix length
if (len != 0) {
len = lps[len - 1];
}
// no prefix matches, set lps[i] to 0
else {
lps[i] = 0;
i++;
}
}
}
return lps;
}
// returns minimum character to be added at
// front to make string palindrome
int minChar(string &s) {
int n = s.length();
string rev = s;
reverse(rev.begin(), rev.end());
// get concatenation of string, special character
// and reverse string
s = s + "$" + rev;
// get LPS array of this concatenated string
vector<int> lps = computeLPSArray(s);
// by subtracting last entry of lps vector from
// string length, we will get our result
return (n - lps.back());
}
int main() {
string s = "aacecaaaa";
cout << minChar(s);
return 0;
}
import java.util.ArrayList;
class GfG {
static int[] computeLPSArray(String pat) {
int n = pat.length();
int[] lps = new int[n];
// lps[0] is always 0
lps[0] = 0;
int len = 0;
// loop calculates lps[i] for i = 1 to n-1
int i = 1;
while (i < n) {
// if the characters match, increment len
// and set lps[i]
if (pat.charAt(i) == pat.charAt(len)) {
len++;
lps[i] = len;
i++;
}
// if there is a mismatch
else {
// if len is not zero, update len to
// the last known prefix length
if (len != 0) {
len = lps[len - 1];
}
// no prefix matches, set lps[i] to 0
else {
lps[i] = 0;
i++;
}
}
}
return lps;
}
// returns minimum character to be added at
// front to make string palindrome
static int minChar(String s) {
int n = s.length();
String rev
= new StringBuilder(s).reverse().toString();
// get concatenation of string, special character
// and reverse string
s = s + "$" + rev;
// get LPS array of this concatenated string
int[] lps = computeLPSArray(s);
// by subtracting last entry of lps array from
// string length, we will get our result
return (n - lps[lps.length - 1]);
}
public static void main(String[] args) {
String s = "aacecaaaa";
System.out.println(minChar(s));
}
}
def computeLPSArray(pat):
n = len(pat)
lps = [0] * n
# lps[0] is always 0
len_lps = 0
# loop calculates lps[i] for i = 1 to n-1
i = 1
while i < n:
# if the characters match, increment len
# and set lps[i]
if pat[i] == pat[len_lps]:
len_lps += 1
lps[i] = len_lps
i += 1
# if there is a mismatch
else:
# if len is not zero, update len to
# the last known prefix length
if len_lps != 0:
len_lps = lps[len_lps - 1]
# no prefix matches, set lps[i] to 0
else:
lps[i] = 0
i += 1
return lps
# returns minimum character to be added at
# front to make string palindrome
def minChar(s):
n = len(s)
rev = s[::-1]
# get concatenation of string, special character
# and reverse string
s = s + "$" + rev
# get LPS array of this concatenated string
lps = computeLPSArray(s)
# by subtracting last entry of lps array from
# string length, we will get our result
return n - lps[-1]
if __name__ == "__main__":
s = "aacecaaaa"
print(minChar(s))
using System;
class GfG {
static int[] computeLPSArray(string pat) {
int n = pat.Length;
int[] lps = new int[n];
// lps[0] is always 0
lps[0] = 0;
int len = 0;
// loop calculates lps[i] for i = 1 to n-1
int i = 1;
while (i < n) {
// if the characters match, increment len
// and set lps[i]
if (pat[i] == pat[len]) {
len++;
lps[i] = len;
i++;
}
// if there is a mismatch
else {
// if len is not zero, update len to
// the last known prefix length
if (len != 0) {
len = lps[len - 1];
}
// no prefix matches, set lps[i] to 0
else {
lps[i] = 0;
i++;
}
}
}
return lps;
}
// minimum character to be added at
// front to make string palindrome
static int minChar(string s) {
int n = s.Length;
char[] charArray = s.ToCharArray();
Array.Reverse(charArray);
string rev = new string(charArray);
// get concatenation of string, special character
// and reverse string
s = s + "$" + rev;
// get LPS array of this concatenated string
int[] lps = computeLPSArray(s);
// by subtracting last entry of lps array from
// string length, we will get our result
return n - lps[lps.Length - 1];
}
static void Main() {
string s = "aacecaaaa";
Console.WriteLine(minChar(s));
}
}
function computeLPSArray(pat) {
let n = pat.length;
let lps = new Array(n).fill(0);
// lps[0] is always 0
let len = 0;
// loop calculates lps[i] for i = 1 to n-1
let i = 1;
while (i < n) {
// if the characters match, increment len
// and set lps[i]
if (pat[i] === pat[len]) {
len++;
lps[i] = len;
i++;
}
// if there is a mismatch
else {
// if len is not zero, update len to
// the last known prefix length
if (len !== 0) {
len = lps[len - 1];
}
// no prefix matches, set lps[i] to 0
else {
lps[i] = 0;
i++;
}
}
}
return lps;
}
// returns minimum character to be added at
// front to make string palindrome
function minChar(s) {
let n = s.length;
let rev = s.split("").reverse().join("");
// get concatenation of string, special character
// and reverse string
s = s + "$" + rev;
// get LPS array of this concatenated string
let lps = computeLPSArray(s);
// by subtracting last entry of lps array from
// string length, we will get our result
return n - lps[lps.length - 1];
}
// Driver Code
let s = "aacecaaaa";
console.log(minChar(s));
[Expected Approach 2] Using Manacher's Algorithm
The idea is to use Manacher’s algorithm to efficiently find all palindromic substrings in linear time.
We transform the string by inserting special characters (#) to handle both even and odd length palindromes uniformly.
After preprocessing, we scan from the end of the original string and use the palindrome radius array to check if the prefix s[0...i] is a palindrome. The first such index i gives us the longest palindromic prefix, and we return n - (i + 1) as the minimum characters to add.
C++
#include <iostream>
#include <vector>
#include <string>
using namespace std;
// manacher's algorithm for finding longest
// palindromic substrings
class manacher {
public:
// array to store palindrome lengths centered
// at each position
vector<int> p;
// modified string with separators and sentinels
string ms;
manacher(string &s) {
ms = "@";
for (char c : s) {
ms += "#" + string(1, c);
}
ms += "#$";
runManacher();
}
// core Manacher's algorithm
void runManacher() {
int n = ms.size();
p.assign(n, 0);
int l = 0, r = 0;
for (int i = 1; i < n - 1; ++i) {
if (i < r)
p[i] = min(r - i, p[r + l - i]);
// expand around the current center
while (ms[i + 1 + p[i]] == ms[i - 1 - p[i]])
++p[i];
// update center if palindrome goes beyond
// current right boundary
if (i + p[i] > r) {
l = i - p[i];
r = i + p[i];
}
}
}
// returns the length of the longest palindrome
// centered at given position
int getLongest(int cen, int odd) {
int pos = 2 * cen + 2 + !odd;
return p[pos];
}
// checks whether substring s[l...r] is a palindrome
bool check(int l, int r) {
int len = r - l + 1;
int longest = getLongest((l + r) / 2, len % 2);
return len <= longest;
}
};
// returns the minimum number of characters to add at the
// front to make the given string a palindrome
int minChar(string &s) {
int n = s.size();
manacher m(s);
// scan from the end to find the longest
// palindromic prefix
for (int i = n - 1; i >= 0; --i) {
if (m.check(0, i))
return n - (i + 1);
}
return n - 1;
}
int main() {
string s = "aacecaaaa";
cout << minChar(s) << endl;
return 0;
}
class GfG {
// manacher's algorithm for finding longest
// palindromic substrings
static class manacher {
// array to store palindrome lengths centered
// at each position
int[] p;
// modified string with separators and sentinels
String ms;
manacher(String s) {
StringBuilder sb = new StringBuilder("@");
for (char c : s.toCharArray()) {
sb.append("#").append(c);
}
sb.append("#$");
ms = sb.toString();
runManacher();
}
// core Manacher's algorithm
void runManacher() {
int n = ms.length();
p = new int[n];
int l = 0, r = 0;
for (int i = 1; i < n - 1; ++i) {
if (i < r)
p[i] = Math.min(r - i, p[r + l - i]);
// expand around the current center
while (ms.charAt(i + 1 + p[i]) == ms.charAt(i - 1 - p[i]))
p[i]++;
// update center if palindrome goes beyond
// current right boundary
if (i + p[i] > r) {
l = i - p[i];
r = i + p[i];
}
}
}
// returns the length of the longest palindrome
// centered at given position
int getLongest(int cen, int odd) {
int pos = 2 * cen + 2 + (odd == 0 ? 1 : 0);
return p[pos];
}
// checks whether substring s[l...r] is a palindrome
boolean check(int l, int r) {
int len = r - l + 1;
int longest = getLongest((l + r) / 2, len % 2);
return len <= longest;
}
}
// returns the minimum number of characters to add at the
// front to make the given string a palindrome
static int minChar(String s) {
int n = s.length();
manacher m = new manacher(s);
// scan from the end to find the longest
// palindromic prefix
for (int i = n - 1; i >= 0; --i) {
if (m.check(0, i))
return n - (i + 1);
}
return n - 1;
}
public static void main(String[] args) {
String s = "aacecaaaa";
System.out.println(minChar(s));
}
}
# manacher's algorithm for finding longest
# palindromic substrings
class manacher:
# array to store palindrome lengths centered
# at each position
def __init__(self, s):
# modified string with separators and sentinels
self.ms = "@"
for c in s:
self.ms += "#" + c
self.ms += "#$"
self.p = []
self.runManacher()
# core Manacher's algorithm
def runManacher(self):
n = len(self.ms)
self.p = [0] * n
l = r = 0
for i in range(1, n - 1):
if i < r:
self.p[i] = min(r - i, self.p[r + l - i])
# expand around the current center
while self.ms[i + 1 + self.p[i]] == self.ms[i - 1 - self.p[i]]:
self.p[i] += 1
# update center if palindrome goes beyond
# current right boundary
if i + self.p[i] > r:
l = i - self.p[i]
r = i + self.p[i]
# returns the length of the longest palindrome
# centered at given position
def getLongest(self, cen, odd):
pos = 2 * cen + 2 + (0 if odd else 1)
return self.p[pos]
# checks whether substring s[l...r] is a palindrome
def check(self, l, r):
length = r - l + 1
longest = self.getLongest((l + r) // 2, length % 2)
return length <= longest
# returns the minimum number of characters to add at the
# front to make the given string a palindrome
def minChar(s):
n = len(s)
m = manacher(s)
# scan from the end to find the longest
# palindromic prefix
for i in range(n - 1, -1, -1):
if m.check(0, i):
return n - (i + 1)
return n - 1
if __name__ == "__main__":
s = "aacecaaaa"
print(minChar(s))
using System;
class GfG {
// manacher's algorithm for finding longest
// palindromic substrings
class manacher {
// array to store palindrome lengths centered
// at each position
public int[] p;
// modified string with separators and sentinels
public string ms;
public manacher(string s) {
ms = "@";
foreach (char c in s) {
ms += "#" + c;
}
ms += "#$";
runManacher();
}
// core Manacher's algorithm
void runManacher() {
int n = ms.Length;
p = new int[n];
int l = 0, r = 0;
for (int i = 1; i < n - 1; ++i) {
if (i < r)
p[i] = Math.Min(r - i, p[r + l - i]);
// expand around the current center
while (ms[i + 1 + p[i]] == ms[i - 1 - p[i]])
p[i]++;
// update center if palindrome goes beyond
// current right boundary
if (i + p[i] > r) {
l = i - p[i];
r = i + p[i];
}
}
}
// returns the length of the longest palindrome
// centered at given position
public int getLongest(int cen, int odd) {
int pos = 2 * cen + 2 + (odd == 0 ? 1 : 0);
return p[pos];
}
// checks whether substring s[l...r] is a palindrome
public bool check(int l, int r) {
int len = r - l + 1;
int longest = getLongest((l + r) / 2, len % 2);
return len <= longest;
}
}
// returns the minimum number of characters to add at the
// front to make the given string a palindrome
static int minChar(string s) {
int n = s.Length;
manacher m = new manacher(s);
// scan from the end to find the longest
// palindromic prefix
for (int i = n - 1; i >= 0; --i) {
if (m.check(0, i))
return n - (i + 1);
}
return n - 1;
}
static void Main() {
string s = "aacecaaaa";
Console.WriteLine(minChar(s));
}
}
// manacher's algorithm for finding longest
// palindromic substrings
class manacher {
// array to store palindrome lengths centered
// at each position
constructor(s) {
// modified string with separators and sentinels
this.ms = "@";
for (let c of s) {
this.ms += "#" + c;
}
this.ms += "#$";
this.p = [];
this.runManacher();
}
// core Manacher's algorithm
runManacher() {
const n = this.ms.length;
this.p = new Array(n).fill(0);
let l = 0, r = 0;
for (let i = 1; i < n - 1; ++i) {
if (i < r)
this.p[i] = Math.min(r - i, this.p[r + l - i]);
// expand around the current center
while (this.ms[i + 1 + this.p[i]] === this.ms[i - 1 - this.p[i]])
this.p[i]++;
// update center if palindrome goes beyond
// current right boundary
if (i + this.p[i] > r) {
l = i - this.p[i];
r = i + this.p[i];
}
}
}
// returns the length of the longest palindrome
// centered at given position
getLongest(cen, odd) {
const pos = 2 * cen + 2 + (odd === 0 ? 1 : 0);
return this.p[pos];
}
// checks whether substring s[l...r] is a palindrome
check(l, r) {
const len = r - l + 1;
const longest = this.getLongest(Math.floor((l + r) / 2), len % 2);
return len <= longest;
}
}
// returns the minimum number of characters to add at the
// front to make the given string a palindrome
function minChar(s) {
const n = s.length;
const m = new manacher(s);
// scan from the end to find the longest
// palindromic prefix
for (let i = n - 1; i >= 0; --i) {
if (m.check(0, i))
return n - (i + 1);
}
return n - 1;
}
// Driver Code
const s = "aacecaaaa";
console.log(minChar(s));
Time Complexity: O(n), manacher's algorithm runs in linear time by expanding palindromes at each center without revisiting characters, and the prefix check loop performs O(1) operations per character over n characters.
Auxiliary Space: O(n), used for the modified string and the palindrome length array p[], both of which grow linearly with the input size.
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Trie Data StructureThe Trie data structure is a tree-like structure used for storing a dynamic set of strings. It allows for efficient retrieval and storage of keys, making it highly effective in handling large datasets. Trie supports operations such as insertion, search, deletion of keys, and prefix searches. In this
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Algorithms
Searching AlgorithmsSearching algorithms are essential tools in computer science used to locate specific items within a collection of data. In this tutorial, we are mainly going to focus upon searching in an array. When we search an item in an array, there are two most common algorithms used based on the type of input
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Sorting AlgorithmsA Sorting Algorithm is used to rearrange a given array or list of elements in an order. For example, a given array [10, 20, 5, 2] becomes [2, 5, 10, 20] after sorting in increasing order and becomes [20, 10, 5, 2] after sorting in decreasing order. There exist different sorting algorithms for differ
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Introduction to RecursionThe process in which a function calls itself directly or indirectly is called recursion and the corresponding function is called a recursive function. A recursive algorithm takes one step toward solution and then recursively call itself to further move. The algorithm stops once we reach the solution
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Greedy AlgorithmsGreedy algorithms are a class of algorithms that make locally optimal choices at each step with the hope of finding a global optimum solution. At every step of the algorithm, we make a choice that looks the best at the moment. To make the choice, we sometimes sort the array so that we can always get
3 min read
Graph AlgorithmsGraph is a non-linear data structure like tree data structure. The limitation of tree is, it can only represent hierarchical data. For situations where nodes or vertices are randomly connected with each other other, we use Graph. Example situations where we use graph data structure are, a social net
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Dynamic Programming or DPDynamic Programming is an algorithmic technique with the following properties.It is mainly an optimization over plain recursion. Wherever we see a recursive solution that has repeated calls for the same inputs, we can optimize it using Dynamic Programming. The idea is to simply store the results of
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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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