# Count of positions j in Array such that arr[i] is maximum in index range [i, j] with end points as same

• Difficulty Level : Medium
• Last Updated : 17 Jul, 2021

Given an array arr[] consisting of N positive integers, the task is to find all j such that arr[j] = arr[i] and all the numbers in the range [min(j, i), max(j, i)] is less than or equal to arr[i] where 1 ≤ i ≤ N.

Examples:

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Input: arr[] = {4, 7, 7, 9, 7}, N = 5
Output: 1 2 2 1 1
Explanation:
For i = 1, j = 1 is the only element such that arr[i] = arr[j] and no element in the middle has value greater than arr.
For i = 2, j = 2 and 3 are the elements such that arr[i] = arr[j] and no elements in the middle has value greater than arr.
For i = 3, j = 2 and 3 are the elements such that arr[i] = arr[j] and no elements in the middle has value greater than arr.
For i = 4, j = 4 is the only element such that arr[i] = arr[j] and no element in the middle has value greater than arr.
For i = 5, j = 5 is the only element such that arr[i] = arr[j] and no element in the middle has value greater than arr.

Input: arr[] = {1, 2, 1, 2, 4}, N = 5
Output: 1 2 1 2 1

Approach: The simplest way to solve the problem is to use two nested for loops to traverse the array and find the pairs such that arr[i] = arr[j] and no element in the range [i, j] is greater than arr[i]. Follow the steps below to solve the problem:

• Initialize an array, say ans[] that stores the answer for all the elements in the range [0, N-1].
• Iterate in the range[N-1, 0] using the variable i and perform the following steps:
• Iterate in the range[i, 0] using the variable j and perform the following steps:
• If arr[j] = arr[i], then increment the value of ans[i] by 1.
• If arr[j] > arr[i], then terminate the loop.
• Iterate in the range[i+1, N-1] using the variable j and perform the following steps:
• If arr[j] = arr[i], then increment the value of ans[i] by 1.
• If arr[j] > arr[i], then terminate the loop.
• After completing the above steps, print the array ans[] as the answer.

Below is the implementation of the above approach

## C++

 `// C++ program for the above approach``#include ``using` `namespace` `std;` `// Function to find j such that arr[j]=arr[i]``// and no element is greater than arr[i] in the``// range [j, i]``void` `findElements(``int` `arr[], ``int` `N)``{` `    ``// To Store answer``    ``int` `ans[N];` `    ``// Initialize the value of ans[i] as 0``    ``for` `(``int` `i = 0; i < N; i++) {``        ``ans[i] = 0;``    ``}` `    ``// Traverse the array in reverse direction``    ``for` `(``int` `i = N - 1; i >= 0; i--) {` `        ``// Traverse in the range [i, 0]``        ``for` `(``int` `j = i; j >= 0; j--) {``            ``if` `(arr[j] == arr[i]) {` `                ``// Increment ans[i] if arr[j] =arr[i]``                ``ans[i]++;``            ``}``            ``else``                ``// Terminate the loop``                ``if` `(arr[j] > arr[i])``                ``break``;``        ``}` `        ``// Traverse in the range [i+1, N-1]``        ``for` `(``int` `j = i + 1; j < N; j++) {``            ``// Increment ans[i] if arr[i] = arr[j]``            ``if` `(arr[j] == arr[i])``                ``ans[i]++;``            ``else` `if` `(arr[j] > arr[i]) {``                ``break``;``            ``}``        ``}``    ``}` `    ``for` `(``int` `i = 0; i < N; i++) {``        ``cout << ans[i] << ``" "``;``    ``}``}` `// Driver Code``int` `main()``{``    ``// Given Input``    ``int` `arr[] = { 1, 2, 1, 2, 4 };``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr);` `    ``// Function Call``    ``findElements(arr, N);``}`

## Java

 `// Java program for the above approach``public` `class` `GFG``{` `    ``// Function to find j such that arr[j]=arr[i]``    ``// and no element is greater than arr[i] in the``    ``// range [j, i]``    ``static` `void` `findElements(``int` `arr[], ``int` `N)``    ``{` `        ``// To Store answer``        ``int` `ans[] = ``new` `int``[N];` `        ``// Initialize the value of ans[i] as 0``        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``ans[i] = ``0``;``        ``}` `        ``// Traverse the array in reverse direction``        ``for` `(``int` `i = N - ``1``; i >= ``0``; i--) {` `            ``// Traverse in the range [i, 0]``            ``for` `(``int` `j = i; j >= ``0``; j--) {``                ``if` `(arr[j] == arr[i]) {` `                    ``// Increment ans[i] if arr[j] =arr[i]``                    ``ans[i]++;``                ``}``                ``else``                    ``// Terminate the loop``                    ``if` `(arr[j] > arr[i])``                    ``break``;``            ``}` `            ``// Traverse in the range [i+1, N-1]``            ``for` `(``int` `j = i + ``1``; j < N; j++)``            ``{``                ``// Increment ans[i] if arr[i] = arr[j]``                ``if` `(arr[j] == arr[i])``                    ``ans[i]++;``                ``else` `if` `(arr[j] > arr[i]) {``                    ``break``;``                ``}``            ``}``        ``}` `        ``for` `(``int` `i = ``0``; i < N; i++) {``            ``System.out.print(ans[i] + ``" "``);``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``// Given Input``        ``int` `arr[] = { ``1``, ``2``, ``1``, ``2``, ``4` `};``        ``int` `N = arr.length;` `        ``// Function Call``        ``findElements(arr, N);``    ``}``}` `// This code is contributed by abhinavjain194`

## Python3

 `# Python3 program for the above approach` `# Function to find j such that arr[j]=arr[i]``# and no element is greater than arr[i] in the``# range [j, i]``def` `findElements(arr, N):``    ` `    ``# Initialising a list to store ans``    ``# Initialize the value of ans[i] as 0``    ``ans ``=` `[``0` `for` `i ``in` `range``(N)]` `    ``# Traverse the array in reverse direction``    ``for` `i ``in` `range``(N ``-` `1``, ``-``1``, ``-``1``):``        ` `        ``# Traverse in the range [i, 0]``        ``for` `j ``in` `range``(i, ``-``1``, ``-``1``):``            ``if` `arr[j] ``=``=` `arr[i]:``                ` `                ``# Increment ans[i] if arr[j] = arr[i]``                ``ans[i] ``+``=` `1``                ` `            ``else``:``                ` `                ``# Terminate the loop``                ``if` `arr[j] > arr[i]:``                    ``break``                    ` `        ``# Traverse in the range [i+1, N-1]``        ``for` `j ``in` `range``(i``+``1``, N):``          ` `            ``# Increment ans[i] if arr[j] = arr[i]``            ``if` `arr[j] ``=``=` `arr[i]:``                ``ans[i] ``+``=` `1``            ``else``:``                ``if` `arr[j] > arr[i]:``                    ``break``                    ` `    ``# Print the ans``    ``for` `i ``in` `range``(N):``        ``print``(ans[i], end ``=` `" "``)``        ` `    ``return` `# Driver Code``if` `__name__ ``=``=` `'__main__'``:``    ` `    ``# Given Input``    ``arr ``=` `[ ``1``, ``2``, ``1``, ``2``, ``4` `]``    ``N ``=` `len``(arr)``    ` `    ``# Function call``    ``findElements(arr, N)` `# This code is contributed by MuskanKalra1`

## C#

 `// C# program for the above approach``using` `System;``using` `System.Collections.Generic;` `class` `GFG{` `// Function to find j such that arr[j]=arr[i]``// and no element is greater than arr[i] in the``// range [j, i]``static` `void` `findElements(``int` `[]arr, ``int` `N)``{``    ` `    ``// To Store answer``    ``int` `[]ans = ``new` `int``[N];` `    ``// Initialize the value of ans[i] as 0``    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ``ans[i] = 0;``    ``}` `    ``// Traverse the array in reverse direction``    ``for``(``int` `i = N - 1; i >= 0; i--)``    ``{` `        ``// Traverse in the range [i, 0]``        ``for``(``int` `j = i; j >= 0; j--)``        ``{``            ``if` `(arr[j] == arr[i])``            ``{``                ` `                ``// Increment ans[i] if arr[j] =arr[i]``                ``ans[i]++;``            ``}``            ``else``            ` `                ``// Terminate the loop``                ``if` `(arr[j] > arr[i])``                    ``break``;``        ``}` `        ``// Traverse in the range [i+1, N-1]``        ``for``(``int` `j = i + 1; j < N; j++)``        ``{``            ` `            ``// Increment ans[i] if arr[i] = arr[j]``            ``if` `(arr[j] == arr[i])``                ``ans[i]++;``            ``else` `if` `(arr[j] > arr[i])``            ``{``                ``break``;``            ``}``        ``}``    ``}` `    ``for``(``int` `i = 0; i < N; i++)``    ``{``        ``Console.Write(ans[i] + ``" "``);``    ``}``}` `// Driver Code``public` `static` `void` `Main()``{``    ` `    ``// Given Input``    ``int` `[]arr = { 1, 2, 1, 2, 4 };``    ``int` `N = arr.Length;` `    ``// Function Call``    ``findElements(arr, N);``}``}` `// This code is contributed by SURENDRA_GANGWAR`

## Javascript

 ``
Output
`1 2 1 2 1 `

Time Complexity: O(N2)
Auxiliary Space: O(N)

My Personal Notes arrow_drop_up