DP28 求能变成回文的最小插入字符个数 Minimum insertions to form a palindrome @geeksforgeeks...
Given a string, find the minimum number of characters to be inserted to convert it to palindrome.
Before we go further, let us understand with few examples:
ab: Number of insertions required is 1.bab
aa: Number of insertions required is 0. aa
abcd: Number of insertions required is 3.dcbabcd
abcda: Number of insertions required is 2. adcbcda which is same as number of insertions in the substring bcd(Why?).
abcde: Number of insertions required is 4.edcbabcde
Let the input string bestr[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 Solution
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
Dynamic Programming based Solution
If we observe the above approach carefully, we can find that it exhibitsoverlapping 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 to be terminated and the recursion tree cannot originate from there. Substring in the same color indicatesoverlapping subproblems.
How to reuse solutions of subproblems?
We can create a table to store results of subproblems so that they can be used directly if 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 diagonal fashion. For the string abcde, 0….4, the following should be order 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)
Another Dynamic Programming Solution (Variation ofLongest Common Subsequence Problem)
The problem of finding minimum insertions can also be solved using Longest Common Subsequence (LCS) Problem. If we find out LCS of string and its reverse, we know how many maximum characters can form a palindrome. We need insert remaining characters. Following are the steps.
1) Find the length of LCS of input string and its reverse. Let the length be ‘l’.
2) The minimum number insertions needed is length of input string minus ‘l’.
package DP;
public class MinInsertionsPalindrome {
public static void main(String[] args) {
String s = "geeks";
int l = 0;
int h = s.length()-1;
System.out.println(findMinInsertionsRec(s, l, h));
System.out.println(findMinInsertionsDP(s, s.length()));
System.out.println(findMinInsertionsLCS(s, s.length()));
}
public static int findMinInsertionsRec(String s, int l, int h){
if(l > h){
return Integer.MAX_VALUE;
}
if(l == h){
return 0;
}
if(l+1 == h){
return (s.charAt(l)==s.charAt(h) ? 0 : 1);
}
if(s.charAt(l) == s.charAt(h)){
return findMinInsertionsRec(s, l+1, h-1);
}else{
return Math.min(findMinInsertionsRec(s, l+1, h), findMinInsertionsRec(s, l, h-1)) + 1;
}
}
// Time complexity: O(N^2) Auxiliary Space: O(N^2)
public static int findMinInsertionsDP(String s, int n){
int[][] dp = new int[n][n];
for(int gap=1; gap