Given two integers L and R where L ≤ R, the task is to find an integer K such that:
- L ≤ K ≤ R.
- All the digits of K are distinct.
- 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: 8Input: 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++
// 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
// 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
# 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#
// 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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