Given a string containing characters as integers only. We need to delete all character of this string in a minimum number of steps where in 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 over all occurrence of the current character in 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 to this subproblem (i+1, K-1) because we can just delete the same character and call for mid substring. We need to take care of a case when first two characters are same in that case we can directly reduce to the subproblem (i+2, j)
So after above discussion of 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] )
Total time complexity of 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 methods
int main()
{
string str = "2553432";
cout << minStepToDeleteString(str) << endl;
return 0;
}
Java
// Java program to find minimum step to
// delete a string
public 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
C#
// C# program to find minimum step to
// delete a string
using 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
Output:
2
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Improved By : nitin mittal
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