We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. Algorithms
  3. Strings
  4. Gemstones
  5. Discussions

Gemstones

Sort by

recency

|

1037 Discussions

|

  • alban_tyrex
     + 0 comments

    Here are my c++ approaches of solving this problem, video explanation here : https://youtu.be/EjhdJXN3a3c

    Solution 1 : Using a list of all possible characters

    int gemstones(vector<string> arr) {
    int result = 0;
    string liste = "abcdefghijklmnopqrstuvwxyz";
    for(int j = 0; j < liste.size(); j++){
        int cp = 0;
        for(int i = 0; i < arr.size(); i++){
            if(arr[i].find(liste[j]) != string::npos) cp++;
        }
        if(cp == arr.size())result++;
    }
    return result;
}

    Solution 2 : Using set of the shortest element of the array of string

    string getShortest(vector<string> arr) {
    string result = arr[0];
    for(int i = 1; i < arr.size(); i++){
        if(arr[i].size() < result.size()) result = arr[i];
    }
    return result;
}
int gemstones(vector<string> arr) {
    int result = 0;
    string shortest = getShortest(arr);
    set<char> liste(shortest.begin(), shortest.end());
    for(auto j = liste.begin(); j != liste.end(); j++){
        int cp = 0;
        for(int i = 0; i < arr.size(); i++){
            if(arr[i].find(*j) != string::npos) cp++;
        }
        if(cp == arr.size())result++;
    }
    return result;
}

  • gunjansahu606
     + 0 comments
    def gemstones(arr):
    # Write your code here
    # Create a set of characters from the first rock
    gemstones_set = set(arr[0])
    
    # Iterate through the remaining rocks
    for i in range(1, len(arr)):
        # Update the gemstones set with the intersection of current rock's characters
        gemstones_set = gemstones_set.intersection(set(arr[i]))
    
    # Return the number of gemstones (size of the set)
    return len(gemstones_set)

  • patrickgao2020
     + 0 comments

    Here is my Python solution!

    def gemstones(arr):
    gems = 0
    rocks = len(arr)
    minerals = set(list("".join(arr)))
    arr = [list(rock) for rock in arr]
    for mineral in minerals:
        if len([True for rock in arr if mineral in rock]) == rocks:
            gems += 1
    return gems

  • adrian_alliskair
     + 0 comments
    

    let contadorPiedrasPreciosas = 0
let piedras = []
const rocas = []
for (const rock of arr) {
    for (const p of rock ) {
        if (piedras.includes(p)) {
            continue;
        } else {
            piedras.push(p)
        }
    }
    rocas.push(piedras)
    piedras = []
}
for (const r of rocas) {
    for (const p of r) {
        aparicionPiedras[p] = (aparicionPiedras[p] || 0) + 1 
    }
}
for (const p in aparicionPiedras) {
    if (aparicionPiedras[p] === arr.length) {
        contadorPiedrasPreciosas++;
    }
}
return contadorPiedrasPreciosas``

    `

  • claudelounou75
     + 0 comments
    int gemstones(int arr_count, char** arr) {
    char* ok = NULL;
    int pres = 0, dragon = 0;
    for(char i = 'a';i <= 'z';i++) {
        for(int j = 0;j < arr_count;j++) {
            ok = strchr(arr[j],i);
            if(ok != NULL) {
                pres++;
            }
            ok = NULL;
        }
        if(pres == arr_count) {
            dragon++;
        }
        pres = 0;
    }
    free(ok);
    return dragon;