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 Returns - 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. Constraints 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;
while(i<len){
int j=i;
int sum =0;
while( j<len && s.charAt(i)==s.charAt(j) ){
sum += (s.charAt(i)-'a') +1;
set.add(sum);
j++;
}
i = j;
}
for(int a0 = 0; a0 < n; a0++){
int x = in.nextInt();
if (set.contains(x)){
System.out.println("Yes");
}
else{
System.out.println("No");
}
}
}
}
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));
scanf("%s",s);
int n;
scanf("%d",&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';
if(id!=bef){
bef=id;
cont=0;
}
cont++;
cnt[id]=fmax(cnt[id],cont);
}
for(int a0 = 0; a0 < n; a0++){
int x;
scanf("%d",&x);
// your code goes here
bool ok = false;
for(int c=0;c<26;c++){
int w = c+1;
if(x%w)continue;
if(x/w > cnt[c])continue;
ok=true;
break;
}
puts(ok?"Yes":"No");
}
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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