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:
- Find the minimum number of insertions in the substring str[l+1,…….h].
- Find the minimum number of insertions in the substring str[l…….h-1].
- 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++
#include<bits/stdc++.h>
using namespace std;
int findMinInsertions(char str[],
int l, int h)
{
if (l > h)
return INT_MAX;
if (l == h)
return 0;
if (l == h - 1)
return (str[l] == str[h]) ? 0 : 1;
return (str[l] == str[h])?
findMinInsertions(str, l + 1, h - 1):
(min(findMinInsertions(str, l, h - 1),
findMinInsertions(str, l + 1, h)) + 1);
}
int main()
{
char str[] = "geeks";
cout << findMinInsertions(str, 0,
strlen(str) - 1);
return 0;
}
|
Time Complexity: O(2^n), where n is the length of the input string. This is because for each recursive call, there are two possibilities: either we insert a character at the beginning of the string or at the end of the string. Therefore, the total number of recursive calls made is equal to the number of binary strings of length n, which is 2^n.
Memoisation based approach(Dynamic Programming):
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.
This gave rise to use dynamic programming approach to store the results of subproblems which can be used later. In this apporach we will go with memoised version and in the next one with tabulation version.
Algorithm:
- Define a function named findMinInsertions which takes a character array str, a two-dimensional vector dp, an integer l and an integer h as arguments.
- If l is greater than h, then return INT_MAX as this is an invalid case.
- If l is equal to h, then return 0 as no insertions are needed.
- If l is equal to h-1, then check if the characters at index l and h are same. If yes, then return 0 else return 1. Store the result in the dp[l][h] matrix.
- If the value of dp[l][h] is not equal to -1, then return the stored value.
- Check if the first and last characters of the string str are same. If yes, then call the function findMinInsertions recursively by passing arguments str, dp, l+1, and h-1.
- If the first and last characters of the string str are not same, then call the function findMinInsertions recursively by passing arguments str, dp, l, and h-1 and also call the function recursively by passing arguments str, dp, l+1, and h. The minimum of the two calls is the answer. Add 1 to it, as one insertion is required to make the string palindrome. Store this result in the dp[l][h] matrix.
- Return the result stored in the dp[l][h] matrix.
Below is the implementation of the approach:
C++
#include <bits/stdc++.h>
using namespace std;
int min(int a, int b)
{ return a < b ? a : b; }
int findMinInsertions(char str[], vector<vector<int>> &dp,
int l, int h)
{
if (l > h)
return INT_MAX;
if (l == h)
return 0;
if (l == h - 1)
return dp[l][h] = ((str[l] == str[h]) ?
0 : 1);
if( dp[l][h] != -1 )
return dp[l][h];
return dp[l][h] = (str[l] == str[h])?
findMinInsertions(str, dp, l + 1, h - 1):
(min(findMinInsertions(str, dp, l, h - 1),
findMinInsertions(str, dp, l + 1, h)) + 1);
}
int main()
{
char str[] = "geeks";
int n = strlen(str);
vector<vector<int>> dp(n, vector<int>(n, -1));
printf("%d",
findMinInsertions(str, dp, 0,
strlen(str)-1));
return 0;
}
|
Time complexity: O(N^2) where N is size of input string.
Auxiliary Space: O(N^2) as 2d dp array has been created to store the states. Here, N is size of input string.
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++
#include <bits/stdc++.h>
using namespace std;
int findMinInsertionsDP(char str[], int n)
{
int table[n][n], l, h, gap;
memset(table, 0, sizeof(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 table[0][n - 1];
}
int main()
{
char str[] = "geeks";
cout << findMinInsertionsDP(str,
strlen(str));
return 0;
}
|
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.
- Find the length of LCS of the input string and its reverse. Let the length be ‘l’.
- The minimum number of insertions needed is the length of the input string minus ‘l’.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
int lcs(string X, string Y,
int m, int n)
{
int L[m+1][n+1];
int i, j;
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]);
}
}
return L[m][n];
}
void reverseStr(string& str)
{
int n = str.length();
for (int i = 0; i < n / 2; i++)
swap(str[i], str[n - i - 1]);
}
int findMinInsertionsLCS(string str, int n)
{
string rev = "";
rev = str;
reverseStr(rev);
return (n - lcs(str, rev, n, n));
}
int main()
{
string str = "geeks";
cout << findMinInsertionsLCS(str,
str.length());
return 0;
}
|
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!