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  1. Prepare
  2. Algorithms
  3. Implementation
  4. Modified Kaprekar Numbers
  5. Discussions

Modified Kaprekar Numbers

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  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can what the explanation here : https://youtu.be/MttrQCHGu3w

    int getSum(long a){
    long sq = a * a;
    int digit = (int)log10(a) + 1;
    int result = sq % (int)pow(10, digit);
    int rest = (int)log10(sq) + 1 - digit;
    if(rest > 0) result += stoi(to_string(sq).substr(0, rest));
    return result;
}

void kaprekarNumbers(int p, int q) {
    bool valid_range = false;
    for(int i = p; i <= q; i++){
        int s = getSum(i);
        if(s == i){
           cout << i << " ";
           valid_range = true; 
        }  
    }
    if(!valid_range) cout << "INVALID RANGE";
}

  • Aurelie123
     + 0 comments

    My Javascript solution

        function kaprekarNumbers(p, q) { 

    // Write your code here

let array = [];

//find all numbers between the range of p and q and add them to an array 

for (let range = p; range <= q; range++) {
    array.push(range)
}

//get square numbers of that array 

let squared = array.map(num => num * num);

//create a second array to store the square roots of the square numbers that are equal to the two parts of the split square number (int1 + int2). The split and conversion from string and back to int is done in the for loop below, looping through the 'squared' array

let array2 = [];

for (let i = 0; i < squared.length; i++) {

    let str = squared[i].toString();
    let max = str.length;
    let str1 = str.slice(0, max / 2);
    let str2 = str.slice(max / 2, max);
    let int1 = parseInt(str1);
    let int2 = parseInt(str2);

    if (Math.sqrt(squared[i]) == (int1 + int2) || Math.sqrt(squared[i]) == squared[0]) {
        array2.push(Math.sqrt(squared[i]))
    }


}

if (array2.length == 0) {
    console.log('INVALID RANGE')
}

else {
    console.log(array2.join(' '))
}

    }

  • tiagoiesbick
     + 0 comments

    Python 3

    def kaprekarNumbers(p, q):
    # Write your code here
    kapN = []    
    for i in range(p, q+1, 1): 
        sqr = str(i**2)
        if i == 1:
            kapN.append(str(i))
        elif len(sqr) < 2:
            pass
        else:         
            d = math.ceil(len(sqr) / 2)           
            r = int(sqr[-d:])
            l = int(sqr[:-d])
            if r > 0 and l > 0 and r+l == i:
                kapN.append(str(i))
    if len(kapN) == 0:
        print('INVALID RANGE')
    print(' '.join(kapN))

  • lucianoinso
     + 0 comments

    Python 3 1-liner

    def kaprekarNumbers(p, q):
    if not [print(n, end=" ") for n in range(p, q + 1) if (int(str(n*n)[:-len(str(n))] or 0) + int(str(n*n)[-len(str(n)):])) == n]: print("INVALID RANGE")

    Almost same code but unraveled

    def kaprekarNumbers(p, q):
    found = False
    for n in range(p, q + 1):
        num = str(n*n)
        l = num[:-len(str(n))]
        r = num[-len(str(n)):]
        if int(l or 0) + int(r) == n:
            found = True
            print(n, end=" ")
    if not found: print("INVALID RANGE")
    return

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can what the explanation here : https://youtu.be/MttrQCHGu3w

    int getSum(long a){
    long sq = a * a;
    int digit = (int)log10(a) + 1;
    int result = sq % (int)pow(10, digit);
    int rest = (int)log10(sq) + 1 - digit;
    if(rest > 0) result += stoi(to_string(sq).substr(0, rest));
    return result;
}

void kaprekarNumbers(int p, int q) {
    bool valid_range = false;
    for(int i = p; i <= q; i++){
        int s = getSum(i);
        if(s == i){
           cout << i << " ";
           valid_range = true; 
        }  
    }
    if(!valid_range) cout << "INVALID RANGE";
}
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