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.


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++ 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)
        // 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;




Time Complexity: O(N)

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