Find the most valued alphabet in the String

Given a string str, the task is to find the maximum valued alphabet in str. The value of a particular alphabet is defined as the difference in the indices of its last and the first occurrence. If there are multiple such alphabets then find the lexicographically smallest alphabet.

Examples:

Input: str = “abbba”
Output: a
value(‘a’) = 4 – 0 = 4
value(‘b’) = 3 – 1 = 2



Input: str = “bbb”
Output: b

Approach: The idea is to store the first and the last occurrences of each of the alphabets in two auxiliary arrays say first[] and last[]. Now, these two arrays can be used to find the maximum valued alphabet in the given string.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
const int MAX = 26;
  
// Function to return the maximum
// valued alphabet
char maxAlpha(string str, int len)
{
  
    // To store the first and the last
    // occurrence of all the characters
    int first[MAX], last[MAX];
  
    // Set the first and the last occurrence
    // of all the characters to -1
    for (int i = 0; i < MAX; i++) {
        first[i] = -1;
        last[i] = -1;
    }
  
    // Update the occurrences of the characters
    for (int i = 0; i < len; i++) {
  
        int index = (str[i] - 'a');
  
        // Only set the first occurrence if
        // it hasn't already been set
        if (first[index] == -1)
            first[index] = i;
  
        last[index] = i;
    }
  
    // To store the result
    int ans = -1, maxVal = -1;
  
    // For every alphabet
    for (int i = 0; i < MAX; i++) {
  
        // If current alphabet doesn't appear
        // in the given string
        if (first[i] == -1)
            continue;
  
        // If the current character has
        // the highest value so far
        if ((last[i] - first[i]) > maxVal) {
            maxVal = last[i] - first[i];
            ans = i;
        }
    }
  
    return (char)(ans + 'a');
}
  
// Driver code
int main()
{
    string str = "abbba";
    int len = str.length();
  
    cout << maxAlpha(str, len);
  
    return 0;
}

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Java

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// Java implementation of the approach
class GFG
{
  
static int MAX = 26;
  
// Function to return the maximum
// valued alphabet
static char maxAlpha(String str, int len)
{
  
    // To store the first and the last
    // occurrence of all the characters
    int []first = new int[MAX];
    int []last = new int[MAX];
  
    // Set the first and the last occurrence
    // of all the characters to -1
    for (int i = 0; i < MAX; i++)
    {
        first[i] = -1;
        last[i] = -1;
    }
  
    // Update the occurrences of the characters
    for (int i = 0; i < len; i++) 
    {
  
        int index = (str.charAt(i) - 'a');
  
        // Only set the first occurrence if
        // it hasn't already been set
        if (first[index] == -1)
            first[index] = i;
  
        last[index] = i;
    }
  
    // To store the result
    int ans = -1, maxVal = -1;
  
    // For every alphabet
    for (int i = 0; i < MAX; i++)
    {
  
        // If current alphabet doesn't appear
        // in the given String
        if (first[i] == -1)
            continue;
  
        // If the current character has
        // the highest value so far
        if ((last[i] - first[i]) > maxVal)
        {
            maxVal = last[i] - first[i];
            ans = i;
        }
    }
    return (char)(ans + 'a');
}
  
// Driver code
public static void main(String[] args)
{
    String str = "abbba";
    int len = str.length();
  
    System.out.print(maxAlpha(str, len));
}
}
  
// This code is contributed by 29AjayKumar

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Python

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# Python implementation of the approach
MAX = 26
  
# Function to return the maximum
# valued alphabet
def maxAlpha(str, len):
  
    # To store the first and the last
    # occurrence of all the characters
  
    # Set the first and the last occurrence
    # of all the characters to -1
      
    first = [-1 for x in range(MAX)]
    last = [-1 for x in range(MAX)]
      
    # Update the occurrences of the characters
    for i in range(0,len): 
  
        index = ord(str[i])-97
  
        # Only set the first occurrence if
        # it hasn't already been set
        if (first[index] == -1):
            first[index] = i
  
        last[index] = i
      
    # To store the result
    ans = -1
    maxVal = -1
  
    # For every alphabet
    for i in range(0,MAX):
  
        # If current alphabet doesn't appear
        # in the given string
        if (first[i] == -1):
            continue
  
        # If the current character has
        # the highest value so far
        if ((last[i] - first[i]) > maxVal):
            maxVal = last[i] - first[i];
            ans = i
  
    return chr(ans + 97)
  
# Driver code
str = "abbba"
len = len(str)
  
print(maxAlpha(str, len))
  
# This code is contributed by Sanjit_Prasad

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C#

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// C# implementation of the approach
using System;
  
class GFG
{
static int MAX = 26;
  
// Function to return the maximum
// valued alphabet
static char maxAlpha(String str, int len)
{
  
    // To store the first and the last
    // occurrence of all the characters
    int []first = new int[MAX];
    int []last = new int[MAX];
  
    // Set the first and the last occurrence
    // of all the characters to -1
    for (int i = 0; i < MAX; i++)
    {
        first[i] = -1;
        last[i] = -1;
    }
  
    // Update the occurrences of the characters
    for (int i = 0; i < len; i++) 
    {
  
        int index = (str[i] - 'a');
  
        // Only set the first occurrence if
        // it hasn't already been set
        if (first[index] == -1)
            first[index] = i;
  
        last[index] = i;
    }
  
    // To store the result
    int ans = -1, maxVal = -1;
  
    // For every alphabet
    for (int i = 0; i < MAX; i++)
    {
  
        // If current alphabet doesn't appear
        // in the given String
        if (first[i] == -1)
            continue;
  
        // If the current character has
        // the highest value so far
        if ((last[i] - first[i]) > maxVal)
        {
            maxVal = last[i] - first[i];
            ans = i;
        }
    }
    return (char)(ans + 'a');
}
  
// Driver code
public static void Main(String[] args)
{
    String str = "abbba";
    int len = str.Length;
  
    Console.Write(maxAlpha(str, len));
}
}
  
// This code is contributed by Rajput-Ji

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Output:

a

Time Complexity: O(N)




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