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# Count index pairs which satisfy the given condition

• Last Updated : 24 May, 2022

Given a permutation P of first N natural numbers, the task is to count the index pairs (i, j) such that P[i] + P[j] = max(P[x]) where i ≤ x ≤ j.
Examples:

Input: P[] = {3, 4, 1, 5, 2}
Output:
Only valid index pairs are (0, 4) and (0, 2)
Input: P[] = {1, 3, 2}
Output: 1

Naive approach: We can solve this problem by iterating for all possible pairs (i, j) and each time gets maximum between them. The time complexity of this approach will be O(n3).
Efficient Approach: Fix the maximum element on a segment and iterate on either the elements to the left of it or to the right of it. If the current maximum is x, and the element we found is y then check whether element x-y can form a subsegment with y (i.e. x is the maximum value on the segment between y and x-y). This works in O(n*n)
But if we can precompute the borders of the segments where x is the maximum element and always choose to iterate on the smaller part of the segment then time complexity will reduce to O(nlogn).
Because every element will be processed no more than logn times, if we process it in a segment of size m, the smaller part of it contains no more than m/2 elements ( which we will process later, and the smaller part of this segment contains no more than m/4 elements, and so on..).
Below is the implementation of the above approach:

## C++

 `// CPP implementation of the approach``#include``using` `namespace` `std;` `// Function to return the count of``// required index pairs``int` `Count_Segment(``int` `p[], ``int` `n)``{``    ``// To store the required count``    ``int` `count = 0;` `    ``// Array to store the left elements``    ``// upto which current element is maximum``    ``int` `upto[n + 1];``    ``for``(``int` `i = 0; i < n + 1; i++)``    ``upto[i] = 0;` `    ``// Iterating through the whole permutation``    ``// except first and last element``    ``int` `j = 0,curr = 0;``    ``for` `(``int` `i = 1; i < n + 1; i++)``    ``{` `        ``// If current element can be maximum``        ``// in a subsegment``        ``if` `(p[i] > p[i - 1] and p[i] > p[i + 1])``        ``{` `            ``// Current maximum``            ``curr = p[i];` `            ``// Iterating for smaller values then``            ``// current maximum on left of it``            ``j = i - 1;``            ``while` `(j >= 0 and p[j] < curr)``            ``{``            ``// Storing left borders``                ``// of the current maximum``                ``upto[p[j]]= curr;``                ``j -= 1;``            ``}` `                `  `            ``// Iterating for smaller values then``            ``// current maximum on right of it``            ``j = i + 1;``            ``while` `(j < n and p[j] < curr)``            ``{` `                ``// Condition satisfies``                ``if` `(upto[curr-p[j]] == curr)``                    ``count += 1;``                ``j+= 1;``            ``}``                ` `        ``}``    ``}` `    ``// Return count of subsegments``    ``return` `count;``}``    `  `// Driver Code``int` `main()``{` `    ``int` `p[] = {3, 4, 1, 5, 2};``    ``int` `n = ``sizeof``(p)/``sizeof``(p);``    ``cout << (Count_Segment(p, n));``    ``return` `0;``}``    ` `// This code is contributed by``// Surendra_Gangwar`

## Java

 `// Java implementation of the approach``import` `java.util.*;` `class` `GFG``{` `// Function to return the count of``// required index pairs``static` `int` `Count_Segment(``int` `p[], ``int` `n)``{``    ``// To store the required count``    ``int` `count = ``0``;` `    ``// Array to store the left elements``    ``// upto which current element is maximum``    ``int` `[]upto = ``new` `int``[n + ``1``];``    ``for``(``int` `i = ``0``; i < n + ``1``; i++)``    ``upto[i] = ``0``;` `    ``// Iterating through the whole permutation``    ``// except first and last element``    ``int` `j = ``0``,curr = ``0``;``    ``for` `(``int` `i = ``1``; i < n ; i++)``    ``{` `        ``// If current element can be maximum``        ``// in a subsegment``        ``if` `(p[i] > p[i - ``1``] && p[i] > p[i + ``1``])``        ``{` `            ``// Current maximum``            ``curr = p[i];` `            ``// Iterating for smaller values then``            ``// current maximum on left of it``            ``j = i - ``1``;``            ``while` `(j >= ``0` `&& p[j] < curr)``            ``{``                ``// Storing left borders``                ``// of the current maximum``                ``upto[p[j]]= curr;``                ``j -= ``1``;``            ``}` `                `  `            ``// Iterating for smaller values then``            ``// current maximum on right of it``            ``j = i + ``1``;``            ``while` `(j < n && p[j] < curr)``            ``{` `                ``// Condition satisfies``                ``if` `(upto[curr-p[j]] == curr)``                    ``count += ``1``;``                ``j+= ``1``;``            ``}``                ` `        ``}``    ``}` `    ``// Return count of subsegments``    ``return` `count;``}``    `  `// Driver Code``public` `static` `void` `main(String[] args)``{``    ``int` `p[] = {``3``, ``4``, ``1``, ``5``, ``2``};``    ``int` `n = p.length;``    ``System.out.println(Count_Segment(p, n));``}``}` `/* This code contributed by PrinciRaj1992 */`

## Python

 `# Python3 implementation of the approach` `# Function to return the count of``# required index pairs``def` `Count_Segment(p, n):``    ` `    ``# To store the required count``    ``count ``=` `0` `    ``# Array to store the left elements``    ``# upto which current element is maximum``    ``upto ``=` `[``False``]``*``(n ``+` `1``)` `    ``# Iterating through the whole permutation``    ``# except first and last element``    ``for` `i ``in` `range``(``1``, n``-``1``):` `        ``# If current element can be maximum``        ``# in a subsegment``        ``if` `p[i]>p[i``-``1``] ``and` `p[i]>p[i ``+` `1``]:` `            ``# Current maximum``            ``curr ``=` `p[i]` `            ``# Iterating for smaller values then``            ``# current maximum on left of it``            ``j ``=` `i``-``1``            ``while` `j>``=` `0` `and` `p[j]

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `// Function to return the count of``// required index pairs``static` `int` `Count_Segment(``int` `[]p, ``int` `n)``{``    ``// To store the required count``    ``int` `count = 0;` `    ``// Array to store the left elements``    ``// upto which current element is maximum``    ``int` `[]upto = ``new` `int``[n + 1];``    ``for``(``int` `i = 0; i < n + 1; i++)``    ``upto[i] = 0;` `    ``// Iterating through the whole permutation``    ``// except first and last element``    ``int` `j = 0,curr = 0;``    ``for` `(``int` `i = 1; i < n ; i++)``    ``{` `        ``// If current element can be maximum``        ``// in a subsegment``        ``if` `(p[i] > p[i - 1] && p[i] > p[i + 1])``        ``{` `            ``// Current maximum``            ``curr = p[i];` `            ``// Iterating for smaller values then``            ``// current maximum on left of it``            ``j = i - 1;``            ``while` `(j >= 0 && p[j] < curr)``            ``{``                ``// Storing left borders``                ``// of the current maximum``                ``upto[p[j]]= curr;``                ``j= j - 1;``            ``}` `                `  `            ``// Iterating for smaller values then``            ``// current maximum on right of it``            ``j = i + 1;``            ``while` `(j < n && p[j] < curr)``            ``{` `                ``// Condition satisfies``                ``if` `(upto[curr-p[j]] == curr)``                    ``count += 1;``                ``j= j+ 1;``            ``}``                ` `        ``}``    ``}` `    ``// Return count of subsegments``    ``return` `count;``}``    `  `// Driver Code``static` `public` `void` `Main ()``{``    ``int` `[]p = {3, 4, 1, 5, 2};``    ``int` `n = p.Length;``    ``Console.WriteLine(Count_Segment(p, n));``}``}` `/* This code contributed by ajit*/`

## Javascript

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Output:

`2`

Time Complexity: O(N2), where N represents the size of the given array.
Auxiliary Space: O(N), where N represents the size of the given array.

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