Minimum substring removals required to make all remaining characters of a string same
Last Updated :
18 May, 2021
Given a string str of length N, the task is to find the minimum number of substrings required to be removed to make all the remaining characters of the string same.
Note: The substring to be removed must not contain the last remaining character in the string.
Examples:
Input: str = "ACBDAB"
Output: 2
Explanation:
Removing the substring { str[1], ..., str[3] } modifies str to "AAB"
Removing the substring {str[2] } modifies str to "AA"
Since all characters of str are equal, the required output is 2.
Input: str = "ZBCDEFZ"
Output: 1
Explanation:
Removing the substring { str[1], ..., str[5] } modifies str to "ZZ"
Since all characters of str are equal, the required output is 1.
Approach: The idea is to first remove all the consecutive duplicate characters of the string and count the frequency of each distinct character of the string. Finally, remove all the characters of the string except the character having minimum frequency. Follow the steps below to solve the problem:
Below is the implementation of the above approach:
C++
// C++ program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to count minimum operations
// required to make all characters equal
// by repeatedly removing substring
void minOperationNeeded(string str)
{
// Remove consecutive duplicate
// characters from str
str = string(str.begin(),
unique(str.begin(), str.end()));
// Stores length of the string
int N = str.length();
// Stores frequency of each distinct
// characters of the string str
int res[256] = { 0 };
// Iterate over all the characters
// of the string str
for (int i = 0; i < N; ++i) {
// Update frequency of str[i]
res[str[i]] += 1;
}
// Decrementing the frequency
// of the string str[0]
res[str[0]] -= 1;
// Decrementing the frequency
// of the string str[N - 1]
res[str[N - 1]] -= 1;
// Stores the required count
int ans = INT_MAX;
// Iterate over all characters
// of the string str
for (int i = 0; i < N; ++i) {
// Update ans
ans = min(ans, res[str[i]]);
}
cout << (ans + 1) << endl;
}
// Driver Code
int main()
{
// Given string
string str = "ABCDABCDABCDA";
minOperationNeeded(str);
return 0;
}
// Java program to implement
// the above approach
import java.util.*;
class GFG {
// Function to count minimum operations
// required to make all characters equal
// by repeatedly removing subString
static void minOperationNeeded(char[] str)
{
// Remove consecutive duplicate
// characters from str
str = modstring(str);
// Stores length of the String
int N = str.length;
// Stores frequency of each distinct
// characters of the String str
int res[] = new int[256];
// Iterate over all the characters
// of the String str
for (int i = 0; i < N; ++i) {
// Update frequency of str[i]
res[str[i]] += 1;
}
// Decrementing the frequency
// of the String str[0]
res[str[0]] -= 1;
// Decrementing the frequency
// of the String str[N - 1]
res[str[N - 1]] -= 1;
// Stores the required count
int ans = Integer.MAX_VALUE;
// Iterate over all characters
// of the String str
for (int i = 0; i < N; ++i) {
// Update ans
ans = Math.min(ans, res[str[i]]);
}
System.out.print((ans + 1) + "\n");
}
private static char[] modstring(char[] str) {
String s = "";
boolean b = true;
for (int i = 1; i < str.length; ++i) {
if (str[i - 1] != str[i])
b = true;
if (b) {
s += str[i-1];
b = false;
}
}
return s.toCharArray();
}
// Driver Code
public static void main(String[] args)
{
// Given String
String str = "ABCDABCDABCDA";
minOperationNeeded(str.toCharArray());
}
}
// This code is contributed by Amit Katiyar
# Python3 program to implement
# the above approach
import re, sys
# Function to count minimum operations
# required to make all characters equal
# by repeatedly removing substring
def minOperationNeeded(s):
# Remove consecutive duplicate
# characters from str
d = {}
str = re.sub(r"(.)\1 + ",'', s)
# Stores length of the string
N = len(str)
# Stores frequency of each distinct
# characters of the string str
res = [0 for i in range(256)]
# Iterate over all the characters
# of the string str
for i in range(N):
# Update frequency of str[i]
res[ord(str[i])] += 1
# Decrementing the frequency
# of the string str[0]
res[ord(str[0])] -= 1
# Decrementing the frequency
# of the ord(string ord(str[N - 1]
res[ord(str[N - 1])] -= 1
# Stores the required count
ans = sys.maxsize
# Iterate over all characters
# of the string str
for i in range(N):
# Update ans
ans = min(ans, res[ord(str[i])])
print ((ans + 1))
# Driver Code
if __name__ == '__main__':
# Given string
str = "ABCDABCDABCDA"
minOperationNeeded(str)
# This code is contributed by mohit kumar 29
// C# program for the above approach
using System;
class GFG{
// Function to count minimum operations
// required to make all characters equal
// by repeatedly removing subString
static void minOperationNeeded(char[] str)
{
// Remove consecutive duplicate
// characters from str
str = modstring(str);
// Stores length of the String
int N = str.Length;
// Stores frequency of each distinct
// characters of the String str
int[] res = new int[256];
// Iterate over all the characters
// of the String str
for (int i = 0; i < N; ++i)
{
// Update frequency of str[i]
res[str[i]] += 1;
}
// Decrementing the frequency
// of the String str[0]
res[str[0]] -= 1;
// Decrementing the frequency
// of the String str[N - 1]
res[str[N - 1]] -= 1;
// Stores the required count
int ans = Int32.MaxValue;
// Iterate over all characters
// of the String str
for (int i = 0; i < N; ++i)
{
// Update ans
ans = Math.Min(ans, res[str[i]]);
}
Console.WriteLine((ans + 1) + "\n");
}
private static char[] modstring(char[] str)
{
string s = "";
bool b = true;
for (int i = 1; i < str.Length; ++i)
{
if (str[i - 1] != str[i])
b = true;
if (b)
{
s += str[i - 1];
b = false;
}
}
return s.ToCharArray();
}
// Driver Code
public static void Main()
{
// Given String
string str = "ABCDABCDABCDA";
minOperationNeeded(str.ToCharArray());
}
}
// This code is contributed by code_hunt.
<script>
// JavaScript program for the above approach
// Function to count minimum operations
// required to make all characters equal
// by repeatedly removing subString
function minOperationNeeded(str) {
// Remove consecutive duplicate
// characters from str
str = modstring(str);
// Stores length of the String
var N = str.length;
// Stores frequency of each distinct
// characters of the String str
var res = new Array(256).fill(0);
// Iterate over all the characters
// of the String str
for (var i = 0; i < N; ++i) {
// Update frequency of str[i]
res[str[i].charCodeAt(0)] += 1;
}
// Decrementing the frequency
// of the String str[0]
res[str[0].charCodeAt(0)] -= 1;
// Decrementing the frequency
// of the String str[N - 1]
res[str[N - 1].charCodeAt(0)] -= 1;
// Stores the required count(Max Integer value)
var ans = 2147483647;
// Iterate over all characters
// of the String str
for (var i = 0; i < N; ++i) {
// Update ans
ans = Math.min(ans, res[str[i].charCodeAt(0)]);
}
document.write(ans + 1 + "<br>");
}
function modstring(str) {
var s = "";
var b = true;
for (var i = 1; i < str.length; ++i) {
if (str[i - 1] !== str[i]) b = true;
if (b) {
s += str[i - 1];
b = false;
}
}
return s.split("");
}
// Driver Code
// Given String
var str = "ABCDABCDABCDA";
minOperationNeeded(str.split(""));
</script>
Time Complexity: O(N)
Auxiliary Space: O(256)
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