Weighted Uniform Strings

Problem Statement :

A weighted string is a string of lowercase English letters where each letter has a weight. Character weights are 1 to 26  from  a to z  as shown below:

The weight of a string is the sum of the weights of its characters. For example:

A uniform string consists of a single character repeated zero or more times. For example, ccc and a are uniform strings, but bcb and cd are not.

Function Description

Complete the weightedUniformStrings function in the editor below.

weightedUniformStrings has the following parameter(s):
- string s: a string
- int queries[n]: an array of integers

- string[n]: an array of strings that answer the queries

Input Format

The first line contains a string s, the original string.
The second line contains an integer n, the number of queries.
Each of the next s lines contains an integer queries[i], the weight of a uniform subtring of s that may or may not exist.


1  <=  length of s,n  <=  10^5
1  <=   queries[i]  <=  10^7
s will only contain lowercase English letters, ascii[a-z].

Solution :


                            Solution in C :

In   C++  :

#include <map>
#include <set>
#include <list>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <string>
#include <bitset>
#include <cstdio>
#include <limits>
#include <vector>
#include <climits>
#include <cstring>
#include <cstdlib>
#include <fstream>
#include <numeric>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <unordered_map>

using namespace std;

bool reach[10000010];

int main(){
    string s;
    cin >> s;
    int val = 0;
    for (int i=0; i<s.size(); i++) {
        if (i > 0 && s[i] != s[i-1]) val = 0;
        val += (s[i]-'a'+1);
        reach[val] = true;
    int n;
    cin >> n;
    for(int a0 = 0; a0 < n; a0++){
        int x;
        cin >> x;
        cout << (reach[x] ? "Yes\n" : "No\n");
    return 0;

In  Java  :

import java.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        String s = in.next();
        int len =  s.length();
        int n = in.nextInt();
        Set<Integer> set = new HashSet<Integer>();
        int i=0;
             int j=i;
             int sum =0;
             while( j<len && s.charAt(i)==s.charAt(j) ){
                 sum += (s.charAt(i)-'a') +1;
            i = j;
        for(int a0 = 0; a0 < n; a0++){
            int x = in.nextInt();
            if (set.contains(x)){

In   C :

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main(){
    char* s = (char *)malloc(512000 * sizeof(char));
    int n; 
    int *cnt = (int*)malloc(32 * sizeof(int));
    for(int i=0;i<26;i++)cnt[i]=0;
    int len = strlen(s);
    int bef = 27, cont = 0;
    for(int i=0;i<len;i++){
        int id = s[i]-'a';
    for(int a0 = 0; a0 < n; a0++){
        int x; 
        // your code goes here
        bool ok = false;
        for(int c=0;c<26;c++){
            int w = c+1;
            if(x/w > cnt[c])continue;
    return 0;

In   Python3 :

s = input().strip()
cost = set()
prev = ''
count = 0
for i in s:
    if i != prev:
        prev = i
        count = 0
    count += 1
    cost.add(count * (ord(i) - ord('a') + 1))
for _ in range(int(input())):
    print("Yes" if int(input()) in cost else "No")

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