# Find the last positive element remaining after repeated subtractions of smallest positive element from all Array elements

Given an array **arr[] **consisting of **N** positive integers, the task is to find the last positive array element remaining after repeated subtractions of the smallest positive array element from all array elements.

**Examples:**

Input:arr[] = {3, 5, 4, 7}Output:2Explanation:

Subtract the smallest positive element from the array, i.e. 3, the new array is arr[] = {0, 2, 1, 4}

Subtract the smallest positive element from the array, i.e. 1, the new array is arr[] = {0, 1, 0, 3}

Subtract the smallest positive element from the array, i.e. 1, the new array is arr[] = {0, 0, 0, 2}

The last remaining element is 2.

Input:arr[] ={2, 6, 7}Output:1

**Naive Approach:** The simplest approach to solve the given problem is to traverse the given array **arr[]** and find the smallest positive element in the array and subtract it from all the elements. Perform this operation until there is only one positive element left in the array. After completing the above steps, print the remaining positive array element.

**Time Complexity:** O(N^{2})**Auxiliary Space:** O(1)

**Efficient Approach:** The above approach can be optimized by observing that the last positive element will be the difference between the **largest array element** and the **second-largest array element**. Follow the steps below to solve the problem:

- If
**N = 1**, print the first element of the array. - Otherwise print the difference between the
**largest**and**second-largest**array elements.

Below is the implementation of the above approach:

## C++

`// C++ program for the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to calculate last positive` `// element of the array` `int` `lastPositiveElement(vector<` `int` `> arr)` `{` ` ` `int` `N = arr.size();` ` ` `// Return the first element if N = 1` ` ` `if` `(N == 1)` ` ` `return` `arr[0];` ` ` `// Stores the greatest and the second` ` ` `// greatest element` ` ` `int` `greatest = -1, secondGreatest = -1;` ` ` `// Traverse the array A[]` ` ` `for` `(` `int` `x : arr) {` ` ` `// If current element is greater` ` ` `// than the greatest element` ` ` `if` `(x >= greatest) {` ` ` `secondGreatest = greatest;` ` ` `greatest = x;` ` ` `}` ` ` `// If current element is greater` ` ` `// than second greatest element` ` ` `else` `if` `(x >= secondGreatest) {` ` ` `secondGreatest = x;` ` ` `}` ` ` `}` ` ` `// Return the final answer` ` ` `return` `greatest - secondGreatest;` `}` `// Driver Code` `int` `main()` `{` ` ` `vector<` `int` `> arr = { 3, 5, 4, 7 };` ` ` `cout << lastPositiveElement(arr);` ` ` `return` `0;` `}` |

## Java

`// JAVA program for the above approach` `import` `java.util.*;` `class` `GFG{` `// Function to calculate last positive` `// element of the array` `static` `int` `lastPositiveElement(` `int` `[] arr)` `{` ` ` `int` `N = arr.length;` ` ` `// Return the first element if N = 1` ` ` `if` `(N == ` `1` `)` ` ` `return` `arr[` `0` `];` ` ` `// Stores the greatest and the second` ` ` `// greatest element` ` ` `int` `greatest = -` `1` `, secondGreatest = -` `1` `;` ` ` `// Traverse the array A[]` ` ` `for` `(` `int` `x : arr) {` ` ` `// If current element is greater` ` ` `// than the greatest element` ` ` `if` `(x >= greatest) {` ` ` `secondGreatest = greatest;` ` ` `greatest = x;` ` ` `}` ` ` `// If current element is greater` ` ` `// than second greatest element` ` ` `else` `if` `(x >= secondGreatest) {` ` ` `secondGreatest = x;` ` ` `}` ` ` `}` ` ` `// Return the final answer` ` ` `return` `greatest - secondGreatest;` `}` `// Driver Code` `public` `static` `void` `main(String[] args){` ` ` `int` `[] arr = { ` `3` `, ` `5` `, ` `4` `, ` `7` `};` ` ` `System.out.print(lastPositiveElement(arr));` `}` `}` `// This code is contributed by sanjoy_62.` |

## Python3

`# Python3 program for the above approach` `# Function to calculate last positive` `# element of the array` `def` `lastPositiveElement(arr) :` ` ` `N ` `=` `len` `(arr);` ` ` `# Return the first element if N = 1` ` ` `if` `(N ` `=` `=` `1` `) :` ` ` `return` `arr[` `0` `];` ` ` `# Stores the greatest and the second` ` ` `# greatest element` ` ` `greatest ` `=` `-` `1` `; secondGreatest ` `=` `-` `1` `;` ` ` `# Traverse the array A[]` ` ` `for` `x ` `in` `arr :` ` ` `# If current element is greater` ` ` `# than the greatest element` ` ` `if` `(x >` `=` `greatest) :` ` ` `secondGreatest ` `=` `greatest;` ` ` `greatest ` `=` `x;` ` ` `# If current element is greater` ` ` `# than second greatest element` ` ` `elif` `(x >` `=` `secondGreatest) :` ` ` `secondGreatest ` `=` `x;` ` ` `# Return the final answer` ` ` `return` `greatest ` `-` `secondGreatest;` `# Driver Code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `arr ` `=` `[ ` `3` `, ` `5` `, ` `4` `, ` `7` `];` ` ` `print` `(lastPositiveElement(arr));` ` ` `# This code is contributed by AnkThon` |

## C#

`// C# program for the above approach` `using` `System;` `class` `GFG {` ` ` `// Function to calculate last positive` ` ` `// element of the array` ` ` `static` `int` `lastPositiveElement(` `int` `[] arr)` ` ` `{` ` ` `int` `N = arr.Length;` ` ` `// Return the first element if N = 1` ` ` `if` `(N == 1)` ` ` `return` `arr[0];` ` ` `// Stores the greatest and the second` ` ` `// greatest element` ` ` `int` `greatest = -1, secondGreatest = -1;` ` ` `// Traverse the array A[]` ` ` `for` `(` `int` `x = 0; x < N; x++) {` ` ` `// If current element is greater` ` ` `// than the greatest element` ` ` `if` `(arr[x] >= greatest) {` ` ` `secondGreatest = greatest;` ` ` `greatest = arr[x];` ` ` `}` ` ` `// If current element is greater` ` ` `// than second greatest element` ` ` `else` `if` `(arr[x] >= secondGreatest) {` ` ` `secondGreatest = arr[x];` ` ` `}` ` ` `}` ` ` `// Return the final answer` ` ` `return` `greatest - secondGreatest;` ` ` `}` ` ` `// Driver Code` ` ` `public` `static` `void` `Main()` ` ` `{` ` ` `int` `[] arr = { 3, 5, 4, 7 };` ` ` `Console.Write(lastPositiveElement(arr));` ` ` `}` `}` `// This code is contributed by subhammahato348.` |

## Javascript

`<script>` ` ` `// JavaScript Program to implement` ` ` `// the above approach` ` ` `// Function to calculate last positive` ` ` `// element of the array` ` ` `function` `lastPositiveElement(arr) {` ` ` `let N = arr.length;` ` ` `// Return the first element if N = 1` ` ` `if` `(N == 1)` ` ` `return` `arr[0];` ` ` `// Stores the greatest and the second` ` ` `// greatest element` ` ` `let greatest = -1, secondGreatest = -1;` ` ` `// Traverse the array A[]` ` ` `for` `(let x of arr) {` ` ` `// If current element is greater` ` ` `// than the greatest element` ` ` `if` `(x >= greatest) {` ` ` `secondGreatest = greatest;` ` ` `greatest = x;` ` ` `}` ` ` `// If current element is greater` ` ` `// than second greatest element` ` ` `else` `if` `(x >= secondGreatest) {` ` ` `secondGreatest = x;` ` ` `}` ` ` `}` ` ` `// Return the final answer` ` ` `return` `greatest - secondGreatest;` ` ` `}` ` ` `// Driver Code` ` ` `let arr = [3, 5, 4, 7];` ` ` `document.write(lastPositiveElement(arr));` ` ` `// This code is contributed by Potta Lokesh` ` ` `</script>` |

**Output:**

2

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

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