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  1. Prepare
  2. Algorithms
  3. Strings
  4. Weighted Uniform Strings
  5. Discussions

Weighted Uniform Strings

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788 Discussions

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  • vuhuutuan0
     + 0 comments

    My answer in Typpescript, simple, noted

    function weightedUniformStrings(s: string, queries: number[]): string[] {
    /**
     * idea is calculate the weight of each character and it pairs
     * then compare it with query, any included are 'Yes' orelse 'No'
     * 
     * example:
     * [s]                  'abccddde'
     *                      [a b c cc d dd ddd e]
     * [ws] will be         [1 2 3 6  4 8  12  5]
     * [queries]            [1, 3, 12, 5, 9, 10]
     *                       x  x  x   x  _  _
     * [return]              Y  Y  Y   Y  N  N
     */

    let ws = []
    let cl = null
    let cc = 0
    for (let i = 0; i < s.length; i++) {
        let cd = s.charCodeAt(i) - 97 + 1

        if (!cl || cl != cd) { cl = cd; cc = 0 }
        ws.push(cl * ++cc)
    }

    return queries.map(query => ws.includes(query) ? 'Yes' : 'No')
}

  • pankaj_bodade129
     + 0 comments
    public static List<String> weightedUniformStrings(String s, List<Integer> queries) {
    // Write your code here
    List<Integer> output = new ArrayList<>();
    List<String> weight = new ArrayList<>();
    int n=0;
    for (int i = 0; i < s.length(); i+=n) {
        n=0;
        for (int j = i; j < s.length(); j++) {
            if (s.charAt(i)==s.charAt(j)) {
                n++;
            }else{
                j=s.length();
            }
        }
        for (int j = 1; j < n+1; j++) {
            output.add(j*((int)s.charAt(i)%96));
        }
    }
    for (Integer integer : queries) {
        if (output.contains(integer)) {
            weight.add("Yes");
        }else{
            weight.add("No");
        }
    }
    
    return weight;
    
    

    }

  • hilalaksoy6107
     + 0 comments

    Simple C Solution;

    include

    include

    include

    include

    include

    include

    include

    include

    include

    include

    int main() {

    int n, x, i, j, k;
int a[27] = {0};
int f = 1;
char s[100000];

fgets(s, sizeof(s), stdin);
s[strcspn(s, "\n")] = 0;

for (i = 0; i < 27; ++i) {
    a[i] = 0;
}

for (i = 0; s[i] != '\0'; ++i) {

    if (s[i] == s[i + 1]) {
        ++f;
    } 
    else {

        if (f > a[s[i] - 'a']) {
            a[s[i] - 'a'] = f;
        }
        f = 1;
    }
}

if (f > a[s[i - 1] - 'a']) {
    a[s[i - 1] - 'a'] = f;
}

scanf("%d", &n);
for (k = 0; k < n; ++k) {
    scanf("%d", &x);

    bool found = false;
    for (i = 1; i <= 26; ++i) {

        if (x % i == 0) {

            if (x / i <= a[i - 1]) {
                found = true;
                break;
            }
        }
    }

    if (found) {
        printf("Yes\n");
    } 
    else {
        printf("No\n");
    }
}

return 0;

    }

  • alban_tyrex
     + 0 comments

    Simple C++ Solution, video here : https://youtu.be/xhpzPQBB9ts

    vector<string> weightedUniformStrings(string s, vector<int> queries) {
    // build the set
    map<int, int> mp;
    int last = 0;
    for(int i = 0; i < s.size(); i++){
        if(i != 0 && s[i] == s[i-1]){
            last += s[i] - 'a' + 1;
            mp[last] = 1;
        }
        else{
            last = s[i] - 'a' + 1;
            mp[last] = 1;
        }
    }
    vector<string>result;
    for(int i = 0; i < queries.size(); i++){
        if(mp[queries[i]]) result.push_back("Yes");
        else result.push_back("No");
    }
    return result;
    
}

  • 2k23_cs2312635
     + 3 comments
    public static List<String> weightedUniformStrings(String s, List<Integer> queries) {
    // Write your code here
		// Youtube : The Adarsh 1M
    Set<Integer>hs = new HashSet<>();
        int prev = 0;
        char cho = s.charAt(0);
        // Map<Character, Integer> mp = new HashMap<>();
        for(char ch : s.toCharArray()) {
            int val = (ch-'a') + 1;
            if(ch == cho){
                val = val+prev;
                prev = val;
            }
            else{
                prev = val;
            }
            cho = ch;
            hs.add(val);
            // mp.put(ch, val);
        }
        List<String> ans = new ArrayList<>();
        for(int num : queries) {
            if(hs.contains(num))
                ans.add("Yes");
            else
                ans.add("No");
        }
        return ans;

    }