Pair with given sum and maximum shortest distance from end

Given an array of N integers and an integer K, pick two distinct elements whose sum is K and find the maximum shortest distance of the picked elements from the endpoints.

Examples:

Input : a[] = {2, 4, 3, 2, 1}
        k = 5.
Output :  2
Explanation:
Select the pair(4, 1). 
Shortest distance of 4 from ends = 2
Shortest distance of 1 from ends = 1
Hence, answer is max(2, 1) = 2      

Input : a[] = {2, 4, 1, 9, 5}
        k = 3
Output : 3
Explanation:
Select the pair (2, 1)
Shortest distance of 2 from ends = 1
Shortest distance of 1 from ends = 3
Hence, answer is max(1, 3) = 3. 

Note: The distance of end elements from ends is 1 and not 0.

Naive approach: The approach is to run two loops and in inner loop check if two elements are making a pair with sum k. If yes, then make answer as maximum of the shortest distances of two elements, compare it with the previous pair’s answer and make answer as minimum of these two. When the loop ends we get the desired output.

Efficient Approach: Clearly, Shortest distance is the distance from left end and distance from right end i.e, min(1+i, N-i). Let us denote shortest distance of i-th element as Di. There is another case where an element in the selected pair is repeated then select minimum of all the shortest distances of occurrences of that element. Run a loop and store shortest distance of all the array elements in another array(let it be D[]). Now, we got shortest distances of all the elements.
Run a for loop. If the picked element is x, then the other element should be k-x. Update the ans with max(D[x], D[k-x]) and at every update, select the minimum of previous and present answer. If k-x is not in the array, then D[k-x] will be INFINITE, which will be initialized already.

C++

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// C++ code to find maximum shortest distance 
// from endpoints
#include <bits/stdc++.h>
using namespace std;
  
// function to find maximum shortest distance
int find_maximum(int a[], int n, int k)
{   
    // stores the shortest distance of every 
    // element in original array.
    unordered_map<int, int> b;
      
    for (int i = 0; i < n; i++) {
        int x = a[i];
          
        // shortest distance from ends
        int d = min(1 + i, n - i); 
        if (b.find(x) == b.end())
            b[x] = d;  
  
        else
  
            /* if duplicates are found, b[x] 
            is replaced with minimum of the
            previous and current position's
            shortest distance*/
            b[x] = min(d, b[x]); 
    }
      
    int ans = INT_MAX;
    for (int i = 0; i < n; i++) {
        int x = a[i];
          
        // similar elements ignore them 
        // cause we need distinct elements    
        if (x != k - x && b.find(k - x) != b.end())         
            ans = min(max(b[x], b[k - x]), ans);        
    }
    return ans;
}
  
// driver code
int main()
{
    int a[] = { 3, 5, 8, 6, 7 };
    int K = 11;
    int n = sizeof(a) / sizeof(a[0]);
    cout << find_maximum(a, n, K) << endl;
    return 0;
}

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Python3

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# Python3 code to find maximum shortest 
# distance from endpoints
  
# function to find maximum shortest distance
def find_maximum(a, n, k):
      
    # stores the shortest distance of every 
    # element in original array.
    b = dict()
      
    for i in range(n):
        x = a[i]
          
        # shortest distance from ends
        d = min(1 + i, n - i)
        if x not in b.keys():
            b[x] = d
        else:
  
            # if duplicates are found, b[x] 
            # is replaced with minimum of the
            # previous and current position's
            # shortest distance*/
            b[x] = min(d, b[x])
      
    ans = 10**9
    for i in range(n):
        x = a[i]
          
        # similar elements ignore them 
        # cause we need distinct elements 
        if (x != (k - x) and (k - x) in b.keys()):       
            ans = min(max(b[x], b[k - x]), ans) 
  
    return ans
  
# Driver code
a = [3, 5, 8, 6, 7]
K = 11
n = len(a)
print(find_maximum(a, n, K))
  
# This code is contributed by mohit kumar

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Output:

 2


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