Minimum Insertion Steps to Make a String Palindrome

shriyanshi24jindal

Mar 28th, 2026

Never

Not a member of GistPad yet? Sign Up, it unlocks many cool features!

Java 10.02 KB | None | 0 0

copy raw download clone embed print report

class Solution {
/**
To make a string a palindrome with minimum insertions,
we try to keep the largest palindromic part of the string unchanged and insert characters for the rest.
Minimum Insertions=n−LPS length
Minimum Deletion=n−LPS length
1. Reverse the string and store it in new string
2. Find Longest Common Subsequence (LCS) between s and reverse(s).
3. That LCS is the Longest Palindromic Subsequence.
*/
public int minInsertions(String s) {
String s2 = new StringBuilder(s).reverse().toString();
int n = s.length();
int m = s2.length();
// create a matrix
int[][] t = new int[n+1][m+1];
// initialise the matrix
for(int i=0; i<n+1; i++)
{
for(int j=0; j<m+1; j++)
{
if(i == 0 || j == 0)
t[i][j] = 0;
}
}
// find LCS
for(int i=1; i<n+1; i++)
{
for(int j=1; j<m+1; j++)
{
if(s.charAt(i-1) == s2.charAt(j-1))
{
t[i][j] = 1 + t[i-1][j-1];
}
else
t[i][j] = Math.max(t[i-1][j], t[i][j-1]);
}
}
int lcsLength = t[n][m];
return (n-lcsLength);
}
}
// s= zzazz
// t= zzazz
// LCS: zzazz
// s= mbadm
// t= mdabm
// LCS: mam = 3
// No of deletion = n-lcs = 5-3 = 2
// no of insertion = m - lcs = 5-3 = 2

RAW Paste Data Copied

class Solution {
    /**
    To make a string a palindrome with minimum insertions, 
    we try to keep the largest palindromic part of the string unchanged and insert characters for the rest.
    Minimum Insertions=n−LPS length
    Minimum Deletion=n−LPS length
    1. Reverse the string and store it in new string
    2. Find Longest Common Subsequence (LCS) between s and reverse(s).
    3. That LCS is the Longest Palindromic Subsequence.
    */
    public int minInsertions(String s) {
        String s2 = new StringBuilder(s).reverse().toString();

int n = s.length();
        int m = s2.length();

// create a matrix
        int[][] t = new int[n+1][m+1];

// initialise the matrix
        for(int i=0; i<n+1; i++)
        {
            for(int j=0; j<m+1; j++)
            {
                if(i == 0 || j == 0)
                    t[i][j] = 0;
            }
        }

// find LCS
        for(int i=1; i<n+1; i++)
        {
            for(int j=1; j<m+1; j++)
            {
                if(s.charAt(i-1) == s2.charAt(j-1))
                {
                    t[i][j] = 1 + t[i-1][j-1];
                }
                else
                    t[i][j] = Math.max(t[i-1][j], t[i][j-1]);
            }
        }

int lcsLength = t[n][m];
        return (n-lcsLength);
    }
}

// s= zzazz
// t= zzazz

// LCS: zzazz

// s= mbadm
// t= mdabm
// LCS: mam = 3
// No of deletion = n-lcs = 5-3 = 2
// no of insertion = m - lcs = 5-3 = 2