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

Modified Kaprekar Numbers

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1039 Discussions

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  • akshay_vijay123
     + 0 comments

    Here is my python code (simple brute force):

    def kaprekarNumbers(p, q):
    arr = []
    for i in range(p, q + 1):
        curr = str(i ** 2)
        d1 = curr[:len(curr) // 2] if len(curr) > 1 else "0"
        d2 = curr[len(curr) // 2:] if len(curr) > 1 else curr
        if int(d1) + int(d2) == i:
            arr.append(str(i))
    print(" ".join(arr) if arr != [] else "INVALID RANGE")

  • 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";
}

  • Kavithakumari351
     + 0 comments

    JavaCode

    public static void kaprekarNumbers(int p, int q) {
        List<Integer> list = new ArrayList<>();
        for (int i = p; i <= q; i++) { 
            long sqr = (long) i * i;
            String str = Long.toString(sqr);
            int len = str.length();
            int numLen = Integer.toString(i).length();
            
            String r = str.substring(len - numLen);    
            String l = str.substring(0, len - numLen);  
            if (l.isEmpty()) {
                l= "0";
            }
            int sum = Integer.parseInt(l) + Integer.parseInt(r);
            if (sum == i) {
                list.add(i);
            }
        }
        if (list.isEmpty()) {
            System.out.print("INVALID RANGE");
        } else {
            for (int num : list) {
                System.out.print(num + " ");
            }
        }
    }

  • g_azabache1
     + 0 comments

    Hi, this is my solution in C# 🐿️

        public static void kaprekarNumbers(int p, int q)
    {
        List<int> numbers = new List<int>();
        for (int i = p; i <= q; i++)
        {
            var digits = i.ToString().ToCharArray().Count();
            var square = Math.Pow(i, 2);
            var square_digits = square.ToString().ToCharArray().Count();
            int left = 0;
            int right = 0;
            if (square_digits > 1)
            {
                int digitsForLeft = square_digits - digits;
                left = Convert.ToInt32(square.ToString().Substring(0, digitsForLeft));
                right = Convert.ToInt32(square.ToString().Substring(digitsForLeft));
            }
            else
            {
                right = (int)square;
            }
            if (left + right == i)
                numbers.Add(i);
        }
        Console.WriteLine(numbers.Count > 0 ? string.Join(" ", numbers) : "INVALID RANGE");
    }

  • abhayad268207
     + 0 comments

    Solution of Kaprekar Numbers in Python :

    def kaprekarNumbers(p, q):
    # Write your code here
    li=[]
    for i in range(p , q+1):
        if(i ==1): 
            li.append(i)
                
        elif(i == 2 or i==3 ):
            continue

        elif(i>3):
            num= i*i
            digitsize = len(str(i))
            strnum=str(num)
            leftsize =  len(strnum)-digitsize
            left = int(strnum[:leftsize])
            right= int(strnum[leftsize:])
            add = left + right
            if( add == i ):
                li.append(i)
                                     
    if(len(li) >0):
        for i in range(0,len(li)):
            print(li[i] ,end=" ")
    else:
        print("INVALID RANGE")