Minimum steps to delete a string after repeated deletion of palindrome substrings
Given a string containing characters as integers only. We need to delete all characters of this string in a minimum number of steps wherein one step we can delete the substring which is a palindrome. After deleting a substring remaining parts are concatenated.
Examples:
Input : s = “2553432” Output : 2 We can delete all character of above string in 2 steps, first deleting the substring s[3, 5] “343” and then remaining string completely s[0, 3] “2552” Input : s = “1234” Output : 4 We can delete all character of above string in 4 steps only because each character need to be deleted separately. No substring of length 2 is a palindrome in above string.
We can solve this problem using Dynamic programming. Let dp[i][j] denotes the number of steps it takes to delete the substring s[i, j]. Each character will be deleted alone or as part of some substring so in the first case we will delete the character itself and call subproblem (i+1, j). In the second case, we will iterate the overall occurrence of the current character on the right side, if K is the index of one such occurrence then the problem will reduce to two subproblems (i+1, K – 1) and (K+1, j). We can reach this subproblem (i+1, K-1) because we can just delete the same character and call for the mid substring. We need to take care of a case when the first two characters are the same in that case we can directly reduce to the subproblem (i+2, j).
So after the above discussion of the relation among subproblems, we can write dp relation as follows,
dp[i][j] = min(1 + dp[i+1][j],
dp[i+1][K-1] + dp[K+1][j], where s[i] == s[K]
1 + dp[i+2][j] )The total time complexity of the above solution is O(n^3)
C++
// C++ program to find minimum step to delete a string#include <bits/stdc++.h>using namespace std;/* method returns minimum step for deleting the string, where in one step a palindrome is removed */int minStepToDeleteString(string str){ int N = str.length(); // declare dp array and initialize it with 0s int dp[N + 1][N + 1]; for (int i = 0; i <= N; i++) for (int j = 0; j <= N; j++) dp[i][j] = 0; // loop for substring length we are considering for (int len = 1; len <= N; len++) { // loop with two variables i and j, denoting // starting and ending of substrings for (int i = 0, j = len - 1; j < N; i++, j++) { // If substring length is 1, then 1 step // will be needed if (len == 1) dp[i][j] = 1; else { // delete the ith char individually // and assign result for subproblem (i+1,j) dp[i][j] = 1 + dp[i + 1][j]; // if current and next char are same, // choose min from current and subproblem // (i+2,j) if (str[i] == str[i + 1]) dp[i][j] = min(1 + dp[i + 2][j], dp[i][j]); /* loop over all right characters and suppose Kth char is same as ith character then choose minimum from current and two substring after ignoring ith and Kth char */ for (int K = i + 2; K <= j; K++) if (str[i] == str[K]) dp[i][j] = min(dp[i+1][K-1] + dp[K+1][j], dp[i][j]); } } } /* Uncomment below snippet to print actual dp tablex for (int i = 0; i < N; i++, cout << endl) for (int j = 0; j < N; j++) cout << dp[i][j] << " "; */ return dp[0][N - 1];}// Driver code to test above methodsint main(){ string str = "2553432"; cout << minStepToDeleteString(str) << endl; return 0;} |
Java
// Java program to find minimum step to// delete a stringpublic class GFG{ /* method returns minimum step for deleting the string, where in one step a palindrome is removed */ static int minStepToDeleteString(String str) { int N = str.length(); // declare dp array and initialize it with 0s int[][] dp = new int[N + 1][N + 1]; for (int i = 0; i <= N; i++) for (int j = 0; j <= N; j++) dp[i][j] = 0; // loop for substring length we are considering for (int len = 1; len <= N; len++) { // loop with two variables i and j, denoting // starting and ending of substrings for (int i = 0, j = len - 1; j < N; i++, j++) { // If substring length is 1, then 1 step // will be needed if (len == 1) dp[i][j] = 1; else { // delete the ith char individually // and assign result for // subproblem (i+1,j) dp[i][j] = 1 + dp[i + 1][j]; // if current and next char are same, // choose min from current and // subproblem (i+2, j) if (str.charAt(i) == str.charAt(i + 1)) dp[i][j] = Math.min(1 + dp[i + 2][j], dp[i][j]); /* loop over all right characters and suppose Kth char is same as ith character then choose minimum from current and two substring after ignoring ith and Kth char */ for (int K = i + 2; K <= j; K++) if (str.charAt(i) == str.charAt(K)) dp[i][j] = Math.min( dp[i + 1][K - 1] + dp[K + 1][j], dp[i][j]); } } } /* Uncomment below snippet to print actual dp tablex for (int i = 0; i < N; i++){ System.out.println(); for (int j = 0; j < N; j++) System.out.print(dp[i][j] + " "); } */ return dp[0][N - 1]; } // Driver code to test above methods public static void main(String args[]) { String str = "2553432"; System.out.println(minStepToDeleteString(str)); }}// This code is contributed by Sumit Ghosh |
Python 3
# Python 3 program to find minimum# step to delete a string# method returns minimum step for# deleting the string, where in one# step a palindrome is removeddef minStepToDeleteString(str): N = len(str) # declare dp array and initialize # it with 0s dp = [[0 for x in range(N + 1)] for y in range(N + 1)] # loop for substring length # we are considering for l in range(1, N + 1): # loop with two variables i and j, denoting # starting and ending of substrings i = 0 j = l - 1 while j < N: # If substring length is 1, # then 1 step will be needed if (l == 1): dp[i][j] = 1 else: # delete the ith char individually # and assign result for subproblem (i+1,j) dp[i][j] = 1 + dp[i + 1][j] # if current and next char are # same, choose min from current # and subproblem (i+2,j) if (str[i] == str[i + 1]): dp[i][j] = min(1 + dp[i + 2][j], dp[i][j]) ''' loop over all right characters and suppose Kth char is same as ith character then choose minimum from current and two substring after ignoring ith and Kth char ''' for K in range(i + 2, j + 1): if (str[i] == str[K]): dp[i][j] = min(dp[i + 1][K - 1] + dp[K + 1][j], dp[i][j]) i += 1 j += 1 # Uncomment below snippet to print # actual dp tablex # for (int i = 0; i < N; i++, cout << endl) # for (int j = 0; j < N; j++) # cout << dp[i][j] << " "; return dp[0][N - 1]# Driver Codeif __name__ == "__main__": str = "2553432" print( minStepToDeleteString(str))# This code is contributed by ChitraNayal |
C#
// C# program to find minimum step to// delete a stringusing System;class GFG { /* method returns minimum step for deleting the string, where in one step a palindrome is removed */ static int minStepToDeleteString(string str) { int N = str.Length; // declare dp array and initialize it // with 0s int [,]dp = new int[N + 1,N + 1]; for (int i = 0; i <= N; i++) for (int j = 0; j <= N; j++) dp[i,j] = 0; // loop for substring length we are // considering for (int len = 1; len <= N; len++) { // loop with two variables i and j, // denoting starting and ending of // substrings for (int i = 0, j = len - 1; j < N; i++, j++) { // If substring length is 1, then 1 // step will be needed if (len == 1) dp[i,j] = 1; else { // delete the ith char individually // and assign result for // subproblem (i+1,j) dp[i,j] = 1 + dp[i + 1,j]; // if current and next char are same, // choose min from current and // subproblem (i+2, j) if (str[i] == str[i + 1]) dp[i,j] = Math.Min(1 + dp[i + 2,j], dp[i,j]); /* loop over all right characters and suppose Kth char is same as ith character then choose minimum from current and two substring after ignoring ith and Kth char */ for (int K = i + 2; K <= j; K++) if (str[i] == str[K]) dp[i,j] = Math.Min( dp[i + 1,K - 1] + dp[K + 1,j], dp[i,j]); } } } /* Uncomment below snippet to print actual dp tablex for (int i = 0; i < N; i++){ System.out.println(); for (int j = 0; j < N; j++) System.out.print(dp[i][j] + " "); } */ return dp[0,N - 1]; } // Driver code to test above methods public static void Main() { string str = "2553432"; Console.Write(minStepToDeleteString(str)); }}// This code is contributed by nitin mittal |
PHP
<?php// PHP program to find minimum step// to delete a string// method returns minimum step for// deleting the string, where in one// step a palindrome is removed */function minStepToDeleteString($str){ $N = strlen($str); // declare dp array and initialize // it with 0s $dp[$N + 1][$N + 1] = array(array()); for ($i = 0; $i <= $N; $i++) for ($j = 0; $j <= $N; $j++) $dp[$i][$j] = 0; // loop for substring length we // are considering for ($len = 1; $len <= $N; $len++) { // loop with two variables i and j, denoting // starting and ending of substrings for ($i = 0, $j = $len - 1; $j < $N; $i++, $j++) { // If substring length is 1, then // 1 step will be needed if ($len == 1) $dp[$i][$j] = 1; else { // delete the ith char individually // and assign result for subproblem (i+1,j) $dp[$i][$j] = 1 + $dp[$i + 1][$j]; // if current and next char are same, // choose min from current and subproblem // (i+2,j) if ($str[$i] == $str[$i + 1]) $dp[$i][$j] = min(1 + $dp[$i + 2][$j], $dp[$i][$j]); /* loop over all right characters and suppose Kth char is same as ith character then choose minimum from current and two substring after ignoring ith and Kth char */ for ($K = $i + 2; $K <= $j; $K++) if ($str[$i] == $str[$K]) $dp[$i][$j] = min($dp[$i + 1][$K - 1] + $dp[$K + 1][$j], $dp[$i][$j]); } } } /* Uncomment below snippet to print actual dp tablex for (int i = 0; i < N; i++, cout << endl) for (int j = 0; j < N; j++) cout << dp[i][j] << " "; */ return $dp[0][$N - 1];}// Driver code$str = "2553432";echo minStepToDeleteString($str), "\n";// This code is contributed by ajit.?> |
Javascript
<script>// Java Script program to find minimum step to// delete a string /* method returns minimum step for deleting the string, where in one step a palindrome is removed */ function minStepToDeleteString(str) { let N = str.length; // declare dp array and initialize it with 0s let dp = new Array(N + 1); for (let i = 0; i <= N; i++) { dp[i]=new Array(N+1); for (let j = 0; j <= N; j++) dp[i][j] = 0; } // loop for substring length we are considering for (let len = 1; len <= N; len++) { // loop with two variables i and j, denoting // starting and ending of substrings for (let i = 0, j = len - 1; j < N; i++, j++) { // If substring length is 1, then 1 step // will be needed if (len == 1) dp[i][j] = 1; else { // delete the ith char individually // and assign result for // subproblem (i+1,j) dp[i][j] = 1 + dp[i + 1][j]; // if current and next char are same, // choose min from current and // subproblem (i+2, j) if (str[i] == str[i+1]) dp[i][j] = Math.min(1 + dp[i + 2][j], dp[i][j]); /* loop over all right characters and suppose Kth char is same as ith character then choose minimum from current and two substring after ignoring ith and Kth char */ for (let K = i + 2; K <= j; K++) if (str[i] == str[K]) dp[i][j] = Math.min( dp[i + 1][K - 1] + dp[K + 1][j], dp[i][j]); } } } /* Uncomment below snippet to print actual dp tablex for (int i = 0; i < N; i++){ System.out.println(); for (int j = 0; j < N; j++) System.out.print(dp[i][j] + " "); } */ return dp[0][N - 1]; } // Driver code to test above methods let str = "2553432"; document.write(minStepToDeleteString(str)); // This code is contributed by avanitrachhadiya2155</script> |
Output:
2
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