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Maximize product of absolute index difference with K
  • Last Updated : 05 Aug, 2020

Given an array A[] consisting of N integers, the task is to find the maximum possible value of K, such that K * |i – j| <= min(Ai, Aj), where (0 ? i < j < N).

 

Given expression, k * |i – j| <= min(Ai, Aj) can also be written as k = floor( min(Ai, Aj) / |i – j| )

Examples: 
 

Input: N = 5, A[ ] = {80, 10, 12, 15, 90} 
Output: 20 
Explanation: 
For i = 0 and j = 4, the maximum possible value of K can be obtained. 
Maximum k = min(A[0], A[4]) / |0 – 4| = min(80, 90) / |-4| = 80/4 = 20
Input: N = 5, A[ ] = {10, 5, 12, 15, 8} 
Output: 12 
Explanation: 
For i = 2 and j = 3, the maximum possible value of K can be obtained. 
Maximum k = min(A[2], A[3]) / |2 – 3| = min(12, 15) / |-1| = 12/1 = 12 
 



 

Naive Approach: 
The simplest approach to solve this problem is to generate all possible pairs from the given array, and for each pair, find the value of K and keep updating the maximum K obtained. Finally, print the maximum value of K obtained. 
Time Complexity: O(N2) 
Auxiliary Space: O(1)
Efficient Approach: 
To optimize the above approach, use Two Pointers technique. Follow the steps given below: 
 

  • Initialize three variables i, j and k. Initially set i = 0 and k = 0.
  • Iterate over the array, starting from j = 1 up to j = N-1.
  • Now, for each pair of A[i] and A[j], find min(A[i], A[j]) / ( j – i ). If it is greater than K, then update K.
  • If A[ j ] >= A[i] / ( j – i ), then update pointer i to j, to make sure that among all element up to i, A[i] will result in maximum K with all upcoming A[j]
  • Finally, print the maximum value of K after the array is traversed completely.

Below is the implementation of the above approach:
 

C++

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// C++ program to implement
// the above appraoch
#include <bits/stdc++.h>
using namespace std;
 
// Function returns maximum
// possible value of k
int solve(int A[], int N)
{
 
    // Pointer i make sure that
    // A[i] will result in max k
    int i = 0;
 
    // Stores maximum possible k
    int k = 0;
 
    for (int j = 1; j < N; j++) {
 
        // Possible value of k for
        // current pair (A[i] and A[j])
        int tempK = min(A[i], A[j])
                    / (j - i);
 
        // If current value exceeds k
        if (tempK > k) {
            // Update the value of k
            k = tempK;
        }
 
        // Update pointer i
        if (A[j] >= A[i] / (j - i))
            i = j;
    }
 
    // Return the maxm possible k
    return k;
}
 
// Driver Code
int main()
{
    int A[] = { 10, 5, 12, 15, 8 };
 
    int N = sizeof(A) / sizeof(A[0]);
 
    cout << solve(A, N);
 
    return 0;
}

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Java

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// Java program to implement
// the above appraoch
class GFG{
 
// Function returns maximum
// possible value of k
static int solve(int A[], int N)
{
     
    // Pointer i make sure that
    // A[i] will result in max k
    int i = 0;
 
    // Stores maximum possible k
    int k = 0;
 
    for(int j = 1; j < N; j++)
    {
         
        // Possible value of k for
        // current pair (A[i] and A[j])
        int tempK = Math.min(A[i], A[j]) /
                              (j - i);
 
        // If current value exceeds k
        if (tempK > k)
        {
             
            // Update the value of k
            k = tempK;
        }
 
        // Update pointer i
        if (A[j] >= A[i] / (j - i))
            i = j;
    }
 
    // Return the maxm possible k
    return k;
}
 
// Driver Code
public static void main(String[] args)
{
    int A[] = { 10, 5, 12, 15, 8 };
    int N = A.length;
     
    System.out.println(solve(A, N));
}
}
 
// This code is contributed by rutvik_56

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Python3

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# Python3 program to implement
# the above appraoch
 
# Function returns maximum
# possible value of k
def solve(A, N):
 
    # Pointer i make sure that
    # A[i] will result in max k
    i = 0
 
    # Stores maximum possible k
    k = 0
 
    for j in range(1, N):
 
        # Possible value of k for
        # current pair (A[i] and A[j])
        tempK = (min(A[i], A[j]) // (j - i))
                     
        # If current value exceeds k
        if (tempK > k):
             
            # Update the value of k
            k = tempK
         
        # Update pointer i
        if (A[j] >= A[i] // (j - i)):
            i = j
 
    # Return the maxm possible k
    return k
 
# Driver Code
if __name__ == "__main__":
     
    A = [ 10, 5, 12, 15, 8 ]
    N = len(A);
 
    print(solve(A, N))
 
# This code is contributed by chitranayal

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C#

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// C# program to implement
// the above appraoch
using System;
class GFG{
  
// Function returns maximum
// possible value of k
static int solve(int[] A, int N)
{
      
    // Pointer i make sure that
    // A[i] will result in max k
    int i = 0;
  
    // Stores maximum possible k
    int k = 0;
  
    for(int j = 1; j < N; j++)
    {
          
        // Possible value of k for
        // current pair (A[i] and A[j])
        int tempK = Math.Min(A[i], A[j]) /
                              (j - i);
  
        // If current value exceeds k
        if (tempK > k)
        {
              
            // Update the value of k
            k = tempK;
        }
  
        // Update pointer i
        if (A[j] >= A[i] / (j - i))
            i = j;
    }
  
    // Return the maxm possible k
    return k;
}
  
// Driver Code
public static void Main(string[] args)
{
    int[] A = { 10, 5, 12, 15, 8 };
    int N = A.Length;
      
    Console.Write(solve(A, N));
}
}
  
// This code is contributed by rock_cool

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Output: 

12


 

Time Complexity: O(N) 
Auxiliary Space: O(1)
 

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