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C++ Program To Find Minimum Insertions To Form A Palindrome | DP-28

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  • Last Updated : 23 Dec, 2021

Given string str, the task is to find the minimum number of characters to be inserted to convert it to a palindrome.

Before we go further, let us understand with a few examples: 

  • ab: Number of insertions required is 1 i.e. bab
  • aa: Number of insertions required is 0 i.e. aa
  • abcd: Number of insertions required is 3 i.e. dcbabcd
  • abcda: Number of insertions required is 2 i.e. adcbcda which is the same as the number of insertions in the substring bcd(Why?).
  • abcde: Number of insertions required is 4 i.e. edcbabcde

Let the input string be str[l……h]. The problem can be broken down into three parts:  

  1. Find the minimum number of insertions in the substring str[l+1,…….h].
  2. Find the minimum number of insertions in the substring str[l…….h-1].
  3. Find the minimum number of insertions in the substring str[l+1……h-1].

Recursive Approach: The minimum number of insertions in the string str[l…..h] can be given as:  

  • minInsertions(str[l+1…..h-1]) if str[l] is equal to str[h]
  • min(minInsertions(str[l…..h-1]), minInsertions(str[l+1…..h])) + 1 otherwise

Below is the implementation of the above approach:  

C++




// A Naive recursive program to find minimum 
// number insertions needed to make a string
// palindrome
#include<bits/stdc++.h>
using namespace std;
  
  
// Recursive function to find  
// minimum number of insertions
int findMinInsertions(char str[], 
                      int l, int h)
{
    // Base Cases
    if (l > h) 
        return INT_MAX;
  
    if (l == h) 
        return 0;
  
    if (l == h - 1) 
        return (str[l] == str[h]) ? 0 : 1;
  
    // Check if the first and last characters are
    // same. On the basis of the comparison result, 
    // decide which subrpoblem(s) to call
    return (str[l] == str[h])? 
            findMinInsertions(str, l + 1, h - 1):
            (min(findMinInsertions(str, l, h - 1),
             findMinInsertions(str, l + 1, h)) + 1);
}
  
// Driver code
int main()
{
    char str[] = "geeks";
    cout << findMinInsertions(str, 0, 
                              strlen(str) - 1);
    return 0;
}
// This code is contributed by Akanksha Rai

Output: 

3

Dynamic Programming based Solution 
If we observe the above approach carefully, we can find that it exhibits overlapping subproblems
Suppose we want to find the minimum number of insertions in string “abcde”:  

                      abcde
            /       |      
           /        |        
           bcde         abcd       bcd  <- case 3 is discarded as str[l] != str[h]
       /   |          /   |   
      /    |         /    |    
     cde   bcd  cd   bcd abc bc
   / |   / |  /| / | 
de cd d cd bc c………………….

The substrings in bold show that the recursion is to be terminated and the recursion tree cannot originate from there. Substring in the same color indicates overlapping subproblems.

How to re-use solutions of subproblems? The memorization technique is used to avoid similar subproblem recalls. We can create a table to store the results of subproblems so that they can be used directly if the same subproblem is encountered again.
The below table represents the stored values for the string abcde. 

a b c d e
----------
0 1 2 3 4
0 0 1 2 3 
0 0 0 1 2 
0 0 0 0 1 
0 0 0 0 0

How to fill the table? 
The table should be filled in a diagonal fashion. For the string abcde, 0….4, the following should be ordered in which the table is filled:

Gap = 1: (0, 1) (1, 2) (2, 3) (3, 4)
Gap = 2: (0, 2) (1, 3) (2, 4)

Gap = 3: (0, 3) (1, 4)

Gap = 4: (0, 4)

Below is the implementation of the above approach: 

C++




// A Dynamic Programming based program to find 
// minimum number insertions needed to make a 
// string palindrome 
#include <bits/stdc++.h>
using namespace std;
  
  
// A DP function to find minimum
// number of insertions 
int findMinInsertionsDP(char str[], int n) 
    // Create a table of size n*n. table[i][j] 
    // will store minimum number of insertions 
    // needed to convert str[i..j] to a palindrome. 
    int table[n][n], l, h, gap; 
  
    // Initialize all table entries as 0 
    memset(table, 0, sizeof(table)); 
  
    // Fill the table 
    for (gap = 1; gap < n; ++gap) 
        for (l = 0, h = gap; h < n; ++l, ++h) 
            table[l][h] = (str[l] == str[h])? 
                           table[l + 1][h - 1] : 
                           (min(table[l][h - 1], 
                            table[l + 1][h]) + 1); 
  
    // Return minimum number of insertions
    // for str[0..n-1] 
    return table[0][n - 1]; 
  
// Driver Code
int main() 
    char str[] = "geeks"
    cout << findMinInsertionsDP(str, 
                                strlen(str)); 
    return 0; 
// This is code is contributed by rathbhupendra

Output: 

3

Time complexity: O(N^2) 
Auxiliary Space: O(N^2)

Another Dynamic Programming Solution (Variation of Longest Common Subsequence Problem) 
The problem of finding minimum insertions can also be solved using Longest Common Subsequence (LCS) Problem. If we find out the LCS of string and its reverse, we know how many maximum characters can form a palindrome. We need to insert the remaining characters. Following are the steps. 

  1. Find the length of LCS of the input string and its reverse. Let the length be ‘l’.
  2. The minimum number of insertions needed is the length of the input string minus ‘l’.

Below is the implementation of the above approach:  

C++




// An LCS based program to find minimum number 
// insertions needed to make a string palindrome 
#include <bits/stdc++.h>
using namespace std;
   
// Returns length of LCS for X[0..m-1], 
// Y[0..n-1]. 
int lcs(string X, string Y, 
        int m, int n) 
    int L[m+1][n+1]; 
    int i, j; 
      
    /* Following steps build L[m+1][n+1] in 
       bottom up fashion. Note that L[i][j]  
       contains length of LCS of X[0..i-1] 
       and Y[0..j-1] */
    for (i = 0; i <= m; i++) 
    
        for (j = 0; j <= n; j++) 
        
        if (i == 0 || j == 0) 
            L[i][j] = 0; 
      
        else if (X[i - 1] == Y[j - 1]) 
            L[i][j] = L[i - 1][j - 1] + 1; 
      
        else
            L[i][j] = max(L[i - 1][j], 
                          L[i][j - 1]); 
        
    
      
    /* L[m][n] contains length of LCS for 
       X[0..n-1] and Y[0..m-1] */
    return L[m][n]; 
  
void reverseStr(string& str) 
    int n = str.length(); 
  
    // Swap character starting from two 
    // corners 
    for (int i = 0; i < n / 2; i++) 
        swap(str[i], str[n - i - 1]); 
  
// LCS based function to find minimum 
// number of insertions 
int findMinInsertionsLCS(string str, int n) 
    // Creata another string to store 
    // reverse of 'str' 
    string rev = ""
    rev = str; 
    reverseStr(rev); 
      
    // The output is length of string minus 
    // length of lcs of str and it reverse 
    return (n - lcs(str, rev, n, n)); 
  
// Driver code
int main() 
    string str = "geeks"
    cout << findMinInsertionsLCS(str, 
                                 str.length()); 
    return 0; 
// This code is contributed by rathbhupendra

Output: 

3

Time complexity: O(N^2) 
Auxiliary Space: O(N^2) 

Please refer complete article on Minimum insertions to form a palindrome | DP-28 for more details!


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