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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