Check if frequency of all characters can become same by one removal

Difficulty Level : Easy
Last Updated : 22 Apr, 2021

Given a string which contains lower alphabetic characters, we need to remove at most one character from this string in such a way that frequency of each distinct character becomes same in the string.

Examples:

Input: str = “xyyz”
Output: Yes
We can remove character ’y’ from above
string to make the frequency of each
character same.
Input: str = “xyyzz”
Output: Yes
We can remove character ‘x’ from above
string to make the frequency of each
character same.
Input: str = “xxxxyyzz”
Output: No
It is not possible to make frequency of
each character same just by removing at
most one character from above string.

Approach: The problem can be solved using the concept of hashing. The main thing to observe in this problem is that the position of characters does not matter here so we will count the frequency of characters, if all of them are the same then we are done and there is no need to remove any character to make frequency of characters same Otherwise we can iterate over all characters one by one and decrease their frequency by one, if all frequencies become same then we will flag that it is possible to make character frequency same by at most one removal and if frequencies don’t match then we will increase that frequency again and loop for other characters.

Below is a dry run of the above approach:

Below is the implementation of the above approach:

C++

// C++ program to get same frequency character
// string by removal of at most one char
#include <bits/stdc++.h>
using namespace std;
#define M 26
 
// Utility method to get index of character ch
// in lower alphabet characters
int getIdx(char ch)
{
    return (ch - 'a');
}
 
// Returns true if all non-zero elements
// values are same
bool allSame(int freq[], int N)
{
    int same;
 
    // get first non-zero element
    int i;
    for (i = 0; i < N; i++) {
        if (freq[i] > 0) {
            same = freq[i];
            break;
        }
    }
 
    // check equality of each element with variable same
    for (int j = i + 1; j < N; j++)
        if (freq[j] > 0 && freq[j] != same)
            return false;
 
    return true;
}
 
// Returns true if we can make all character
// frequencies same
bool possibleSameCharFreqByOneRemoval(string str)
{
    int l = str.length();
 
    // fill frequency array
    int freq[M] = { 0 };
    for (int i = 0; i < l; i++)
        freq[getIdx(str[i])]++;
 
    // if all frequencies are same, then return true
    if (allSame(freq, M))
        return true;
 
    /*  Try decreasing frequency of all character
        by one and then    check all equality of all
        non-zero frequencies */
    for (char c = 'a'; c <= 'z'; c++) {
        int i = getIdx(c);
 
        // Check character only if it occurs in str
        if (freq[i] > 0) {
            freq[i]--;
 
            if (allSame(freq, M))
                return true;
            freq[i]++;
        }
    }
 
    return false;
}
 
// Driver code to test above methods
int main()
{
    string str = "xyyzz";
    if (possibleSameCharFreqByOneRemoval(str))
        cout << "Yes";
    else
        cout << "No";
}

Java

// Java program to get same frequency character
// string by removal of at most one char
public class GFG {
 
    static final int M = 26;
 
    // Utility method to get index of character ch
    // in lower alphabet characters
    static int getIdx(char ch)
    {
        return (ch - 'a');
    }
 
    // Returns true if all non-zero elements
    // values are same
    static boolean allSame(int freq[], int N)
    {
        int same = 0;
 
        // get first non-zero element
        int i;
        for (i = 0; i < N; i++) {
            if (freq[i] > 0) {
                same = freq[i];
                break;
            }
        }
 
        // check equality of each element with
        // variable same
        for (int j = i + 1; j < N; j++)
            if (freq[j] > 0 && freq[j] != same)
                return false;
 
        return true;
    }
 
    // Returns true if we can make all character
    // frequencies same
    static boolean possibleSameCharFreqByOneRemoval(String str)
    {
        int l = str.length();
 
        // fill frequency array
        int[] freq = new int[M];
 
        for (int i = 0; i < l; i++)
            freq[getIdx(str.charAt(i))]++;
 
        // if all frequencies are same, then return true
        if (allSame(freq, M))
            return true;
 
        /*  Try decreasing frequency of all character
            by one and then check all equality of all
            non-zero frequencies */
        for (char c = 'a'; c <= 'z'; c++) {
            int i = getIdx(c);
 
            // Check character only if it occurs in str
            if (freq[i] > 0) {
                freq[i]--;
 
                if (allSame(freq, M))
                    return true;
                freq[i]++;
            }
        }
 
        return false;
    }
 
    // Driver code to test above methods
    public static void main(String args[])
    {
        String str = "xyyzz";
        if (possibleSameCharFreqByOneRemoval(str))
            System.out.println("Yes");
        else
            System.out.println("No");
    }
}
// This code is contributed by Sumit Ghosh

Python3

# Python3 program to get same frequency character
# string by removal of at most one char
M = 26
 
# Utility method to get index of character ch
# in lower alphabet characters
def getIdx(ch):
    return (ord(ch) - ord('a'))
 
# Returns true if all non-zero elements
# values are same
def allSame(freq, N):
     
    # get first non-zero element
    for i in range(0, N):
        if(freq[i] > 0):
            same = freq[i]
            break
         
    # check equality of each element
    # with variable same    
    for j in range(i + 1, N):
        if(freq[j] > 0 and freq[j] != same):
            return False
 
    return True
 
# Returns true if we can make all
# character frequencies same
def possibleSameCharFreqByOneRemoval(str1):
    l = len(str1)
 
    # fill frequency array
    freq = [0] * M
    for i in range(0, l):
        freq[getIdx(str1[i])] += 1
         
    # if all frequencies are same,
    # then return true
    if(allSame(freq, M)):
        return True
     
    # Try decreasing frequency of all character
    # by one and then check all equality of all
    # non-zero frequencies
    for i in range(0, 26):
         
        # Check character only if it
        # occurs in str
        if(freq[i] > 0):
            freq[i] -= 1
 
            if(allSame(freq, M)):
                return True
            freq[i] += 1
 
    return False
 
# Driver code
if __name__ == "__main__":
    str1 = "xyyzz"
    if(possibleSameCharFreqByOneRemoval(str1)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by Sairahul099

C#

// C# program to get same frequency
// character string by removal of
// at most one char
using System;
 
class GFG {
    static int M = 26;
 
    // Utility method to get
    // index of character ch
    // in lower alphabet characters
    static int getIdx(char ch)
    {
        return (ch - 'a');
    }
 
    // Returns true if all
    // non-zero elements
    // values are same
    static bool allSame(int[] freq,
                        int N)
    {
        int same = 0;
 
        // get first non-zero element
        int i;
        for (i = 0; i < N; i++) {
            if (freq[i] > 0) {
                same = freq[i];
                break;
            }
        }
 
        // check equality of
        // each element with
        // variable same
        for (int j = i + 1; j < N; j++)
            if (freq[j] > 0 && freq[j] != same)
                return false;
 
        return true;
    }
 
    // Returns true if we
    // can make all character
    // frequencies same
    static bool possibleSameCharFreqByOneRemoval(string str)
    {
        int l = str.Length;
 
        // fill frequency array
        int[] freq = new int[M];
 
        for (int i = 0; i < l; i++)
            freq[getIdx(str[i])]++;
 
        // if all frequencies are same,
        // then return true
        if (allSame(freq, M))
            return true;
 
        /* Try decreasing frequency of all
            character by one and then check
            all equality of all non-zero
            frequencies */
        for (char c = 'a'; c <= 'z'; c++) {
            int i = getIdx(c);
 
            // Check character only if
            // it occurs in str
            if (freq[i] > 0) {
                freq[i]--;
 
                if (allSame(freq, M))
                    return true;
                freq[i]++;
            }
        }
 
        return false;
    }
 
    // Driver code
    public static void Main()
    {
        string str = "xyyzz";
        if (possibleSameCharFreqByOneRemoval(str))
            Console.Write("Yes");
        else
            Console.Write("No");
    }
}
 
// This code is contributed
// by ChitraNayal

PHP

<?php
// PHP program to get same frequency
// character string by removal of at
// most one char
$M = 26;
 
// Utility method to get index
// of character ch in lower
// alphabet characters
function getIdx($ch)
{
    return ($ch - 'a');
}
 
// Returns true if all
// non-zero elements
// values are same
function allSame(&$freq, $N)
{
    // get first non-zero element
    for ($i = 0; $i < $N; $i++)
    {
        if ($freq[$i] > 0)
        {
            $same = $freq[$i];
            break;
        }
    }
 
    // check equality of each
    // element with variable same
    for ($j = $i + 1; $j < $N; $j++)
        if ($freq[$j] > 0 &&
            $freq[$j] != $same)
            return false;
 
    return true;
}
 
// Returns true if we
// can make all character
// frequencies same
function possibleSameCharFreqByOneRemoval($str)
{
    global $M;
    $l = strlen($str);
 
    // fill frequency array
    $freq = array_fill(0, $M, NULL);
    for ($i = 0; $i < $l; $i++)
        $freq[getIdx($str[$i])]++;
 
    // if all frequencies are same,
    // then return true
    if (allSame($freq, $M))
        return true;
 
    /* Try decreasing frequency of all
        character by one and then check
        all equality of all non-zero
        frequencies */
    for ($c = 'a'; $c <= 'z'; $c++)
    {
        $i = getIdx($c);
 
        // Check character only
        // if it occurs in str
        if ($freq[$i] > 0)
        {
            $freq[$i]--;
 
            if (allSame($freq, $M))
                return true;
            $freq[$i]++;
        }
    }
 
    return false;
}
 
// Driver code
$str = "xyyzz";
if (possibleSameCharFreqByOneRemoval($str))
echo "Yes";
else
echo "No";
 
// This code is contributed
// by ChitraNayal
?>

Javascript

<script>
 
// Javascript program to get same
// frequency character string by
// removal of at most one char
let  M = 26;
 
// Utility method to get index of character
// ch in lower alphabet characters
function getIdx(ch)
{
    return (ch - 'a');
}
 
// Returns true if all non-zero elements
// values are same
function allSame(freq, N)
{
    let same = 0;
 
    // Get first non-zero element
    let i;
    for(i = 0; i < N; i++)
    {
        if (freq[i] > 0)
        {
            same = freq[i];
            break;
        }
    }
 
    // Check equality of each element with
    // variable same
    for(let j = i + 1; j < N; j++)
        if (freq[j] > 0 && freq[j] != same)
            return false;
 
    return true;
}
 
// Returns true if we can make all character
// frequencies same
function possibleSameCharFreqByOneRemoval(str)
{
    let l = str.length;
 
    // Fill frequency array
    let freq = new Array(M);
    for(let i = 0; i < M; i++)
    {
        freq[i] = 0;
    }
 
    for(let i = 0; i < l; i++)
        freq[getIdx(str[i])]++;
 
    // If all frequencies are same,
    // then return true
    if (allSame(freq, M))
        return true;
 
    // Try decreasing frequency of all character
    // by one and then check all equality of all
    // non-zero frequencies
    for(let c = 'a'; c <= 'z'; c++)
    {
        let i = getIdx(c);
 
        // Check character only if
        // it occurs in str
        if (freq[i] > 0)
        {
            freq[i]--;
 
            if (allSame(freq, M))
                return true;
                 
            freq[i]++;
        }
    }
    return false;
}
 
// Driver code
let str = "xyyzz";
 
if (possibleSameCharFreqByOneRemoval(str))
    document.write("Yes");
else
    document.write("No");
 
// This code is contributed by avanitrachhadiya2155
 
</script>

Output:

Yes

Time Complexity: O(n) assuming alphabet size is constant.

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

My Personal Notes