Find an integer in the given range that satisfies the given conditions

Given two integers L and R where L ≤ R, the task is to find an integer K such that:

  1. L ≤ K ≤ R.
  2. All the digits of K are distinct.
  3. The value of the expression (L – K) * (K – R) is maximum.

If multiple answers exist then choose the larger value for K.

Examples:



Input: L = 5, R = 10
Output: 8

Input: L = 50, R = 60
Output: 56

Approach: Iterate from L to R and for each value of K, check whether it contains all distinct digits and (L – K) * (K – R) is maximum. If two or more values give the same maximum value for the expression then choose the greater value for K.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
const int MAX = 10;
  
// Function that returns true if x
// contains all distinct digits
bool distinctDigits(int x)
{
    bool present[MAX] = { false };
  
    while (x > 0) {
  
        // Last digit of x
        int digit = x % 10;
  
        // If current digit has
        // appeared before
        if (present[digit])
            return false;
  
        // Mark the current digit
        // to present
        present[digit] = true;
  
        // Remove the last digit
        x /= 10;
    }
  
    return true;
}
  
// Function to return the
// required value of k
int findK(int l, int r)
{
  
    // To store the maximum value
    // for the given expression
    int maxExp = INT_MIN;
    int k = -1;
    for (int i = l; i <= r; i++) {
  
        // If i contains all distinct digits
        if (distinctDigits(i)) {
            int exp = (l - i) * (i - r);
  
            // If the value of the expression
            // is also maximum then update k
            // and the expression
            if (exp >= maxExp) {
                k = i;
                maxExp = exp;
            }
        }
    }
  
    return k;
}
  
// Driver code
int main()
{
    int l = 50, r = 60;
  
    cout << findK(l, r);
  
    return 0;
}

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Java














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// Java implementation of the approach 


class GFG 


{ 


  


static int MAX = 10; 


  


// Function that returns true if x 


// contains all distinct digits 


static boolean distinctDigits(int x) 


{ 


    boolean []present = new boolean[MAX]; 


  


    while (x > 0)  


    { 


  


        // Last digit of x 


        int digit = x % 10; 


  


        // If current digit has 


        // appeared before 


        if (present[digit]) 


            return false; 


  


        // Mark the current digit 


        // to present 


        present[digit] = true; 


  


        // Remove the last digit 


        x /= 10; 


    } 


  


    return true; 


} 


  


// Function to return the 


// required value of k 


static int findK(int l, int r) 


{ 


  


    // To store the maximum value 


    // for the given expression 


    int maxExp = Integer.MIN_VALUE; 


    int k = -1; 


    for (int i = l; i <= r; i++) 


    { 


  


        // If i contains all distinct digits 


        if (distinctDigits(i))  


        { 


            int exp = (l - i) * (i - r); 


  


            // If the value of the expression 


            // is also maximum then update k 


            // and the expression 


            if (exp >= maxExp) 


            { 


                k = i; 


                maxExp = exp; 


            } 


        } 


    } 


  


    return k; 


} 


  


// Driver code 


public static void main(String[] args) 


{ 


    int l = 50, r = 60; 


  


    System.out.print(findK(l, r)); 


  


} 


} 


  


// This code is contributed by 29AjayKumar 



















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Python 3














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# Python3 implementation of the approach 


import sys 


MAX = 10


  


# Function that returns true if x 


# contains all distinct digits 


def distinctDigits(x): 


    present = [False for i in range(MAX)] 


  


    while (x > 0): 


          


        # Last digit of x 


        digit = x % 10


  


        # If current digit has 


        # appeared before 


        if (present[digit]): 


            return False


  


        # Mark the current digit 


        # to present 


        present[digit] = True


  


        # Remove the last digit 


        x = x // 10


  


    return True


  


# Function to return the 


# required value of k 


def findK(l, r): 


      


    # To store the maximum value 


    # for the given expression 


    maxExp = -sys.maxsize - 1


    k = -1


    for i in range(l, r + 1, 1): 


          


        # If i contains all distinct digits 


        if (distinctDigits(i)): 


            exp = (l - i) * (i - r) 


  


            # If the value of the expression 


            # is also maximum then update k 


            # and the expression 


            if (exp >= maxExp): 


                k = i; 


                maxExp = exp 


    return k 


  


# Driver code 


if __name__ == '__main__': 


    l = 50


    r = 60


  


    print(findK(l, r)) 


      


# This code is contributed by Surendra_Gangwar 



















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C#














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// C# implementation of the approach 


using System; 


  


class GFG 


{ 


static int MAX = 10; 


  


// Function that returns true if x 


// contains all distinct digits 


static bool distinctDigits(int x) 


{ 


    bool []present = new bool[MAX]; 


  


    while (x > 0)  


    { 


  


        // Last digit of x 


        int digit = x % 10; 


  


        // If current digit has 


        // appeared before 


        if (present[digit]) 


            return false; 


  


        // Mark the current digit 


        // to present 


        present[digit] = true; 


  


        // Remove the last digit 


        x /= 10; 


    } 


    return true; 


} 


  


// Function to return the 


// required value of k 


static int findK(int l, int r) 


{ 


  


    // To store the maximum value 


    // for the given expression 


    int maxExp = int.MinValue; 


    int k = -1; 


    for (int i = l; i <= r; i++) 


    { 


  


        // If i contains all distinct digits 


        if (distinctDigits(i))  


        { 


            int exp = (l - i) * (i - r); 


  


            // If the value of the expression 


            // is also maximum then update k 


            // and the expression 


            if (exp >= maxExp) 


            { 


                k = i; 


                maxExp = exp; 


            } 


        } 


    } 


    return k; 


} 


  


// Driver code 


public static void Main(String[] args) 


{ 


    int l = 50, r = 60; 


  


    Console.Write(findK(l, r)); 


} 


} 


  


// This code is contributed by Rajput-Ji 



















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Output:


56





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