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  1. Prepare
  2. Algorithms
  3. Strings
  4. How Many Substrings?
  5. Discussions

How Many Substrings?

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70 Discussions

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  • rqadri133
     + 0 comments

    I am having memory abort issues for some test cases

  • irrfankhann29
     + 0 comments

    solution in JS:

    getting timeout error because the time complexity is O(n*m*k), help me to reduce the complexity:

    function countSubstrings(s, queries) {
    // Write your code here
    let left;
    let right;
    queries.forEach((query, index) => {
        let mainSubString = [];
        left = query[0];
        right = query[1];
        const subString = s.slice(left, right+1);
        const sstrLength = subString.length;
        for(let i=0; i<sstrLength; i++){
        for (let j = i + 1; j <= sstrLength; j++) {
            if(!mainSubString.includes(
                subString.slice(i, j))){
            mainSubString.push(subString.slice(i, j))
            }

        }
        }
        console.log(mainSubString.length)  
    })

}

  • yashparihar729
     + 0 comments

    Here is my solution in java, C++, Csharp HackerRank How Many Substrings? Solution

  • mineman1012221
     + 0 comments

    Here are the solution of How Many Substrings? Click Here

  • genewangtp
     + 1 comment

    Sliding window would not help, already saw the solution, maybe it's a math test, but unfortunately I'm not a genius, if you can refactor my code, please help me.. the code only pass some case.

    int counting(string s, int length){
    int re=0;
    unordered_map<string, bool>mp;
    for(int i=0;i<length;i++){
        for(int j=1;j<=length-i;j++){
            string sub = s.substr(i,j);
            if(!mp[sub]){
                mp[sub]=true;
                re++;
            }
        }
    }    
    
    return re;
}
vector<int> countSubstrings(string s, vector<vector<int>> queries) {
    
    int length=queries.size();
    vector<int> re;
    
    for(int i=0;i<length;i++){
        int left = queries[i][0];
        int right = (left > 0) ? queries[i][1] : queries[i][1]+1;
        int length = queries[i][1]-queries[i][0]+1;
        int count = counting(s.substr(left, right), length);
        re.push_back(count);
    }
    return re;
}