Digit Count in Range

C++Server Side ProgrammingProgramming

Suppose we have an integer d between 0 and 9, we also have two positive integers low and high as lower and upper bounds, respectively. We have to find the number of times that d occurs as a digit in all integers between low and high, including the bounds low and high.

So, if the input is like d = 1, low = 1, high = 13, then the output will be 6, as digit d=1 occurs 6 times like 1,10,11,12,13.

To solve this, we will follow these steps −

Define a function zero(), this will take n,

  • ret := 0, x := 0

  • if n is same as 0, then −

    • return 1

  • for initialize m := 1, when m <= n, update m := m * 10, do −

    • a := n / m

    • b := n mod m

    • z := a mod 10

    • if number of digits in m is same as number of digits in n, then −

      • Come out from the loop

    • if z > x, then −

      • ret := ret + ((a / 10) + 1)

    • otherwise when z is same as x, then

      • ret := ret + ((a / 10) * m + (b + 1))

    • Otherwise

      • ret := ret + (a / 10)

  • return ret

  • Define a function f(), this will take x, n,

  • ret := 0

  • for initialize m := 1, when m <= n, update m := m * 10, do −

    • a := n / m

    • b := n mod m

    • z := a mod 10

    • if z > x, then

      • ret := ret + ((a / 10) + 1)

    • otherwise when z is same as x, then −

      • ret := ret + ((a / 10) * m + (b + 1))

    • Otherwise

      • ret := ret + (a / 10)

    • if x is same as 0, then −

      • ret := ret - m

  • return ret

  • From the main method do the following

  • return ret

  • return f(d, high - f(d, low - 1))

Let us see the following implementation to get better understanding −

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
   public:
   int digitCount(int x){
      int ret = 0;
      while (x) {
         ret++;
         x /= 10;
      }
      return ret;
   }
   int zero(int n){
      int ret = 0;
      int x = 0;
      if (n == 0)
      return 1;
      for (int m = 1; m <= n; m *= 10) {
         int a = n / m;
         int b = n % m;
         int z = a % 10;
         if (digitCount(m) == digitCount(n))
         break;
         if (z > x) {
            ret += ((a / 10) + 1) * m;
         } 
         else if (z == x) {
            ret += (a / 10) * m + (b + 1);
         } else {
            ret += (a / 10) * m;
         }
         cout << ret << endl;
      }
      return ret;
   }
   int f(int x, int n){
      int ret = 0;
      for (int m = 1; m <= n; m *= 10) {
         int a = n / m;
         int b = n % m;
         int z = a % 10;
         if (z > x) {
            ret += ((a / 10) + 1) * m;
         }
         else if (z == x) {
            ret += (a / 10) * m + (b + 1);
         } else {
            ret += (a / 10) * m;
         }
         if (x == 0) {
            ret -= m;
         }
      }
      return ret;
   }
   int digitsCount(int d, int low, int high){
      return f(d, high) - f(d, low - 1);
   }
};
main(){
   Solution ob;
   cout << (ob.digitsCount(1,1,13));
}

Input

1,1,13

Output

6
raja
Published on 11-Jul-2020 12:02:50
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