# Find the resulting output array after doing given operations

• Difficulty Level : Expert
• Last Updated : 21 Jun, 2022

Given an array integers[] of integers of size N. Find the resulting output array after doing some operations:

• If the (i-1)th element is positive and ith element is negative  then :
• If the absolute of (i)th is greater than (i-t)th, remove all the contiguous positive elements from left having absolute value less than ith.
• If the absolute of (i)th is smaller than (i-t)th, remove this element.
• If the absolute of (i)th is equal to (i-1)th then remove both the elements.

Examples:

Input: integers[] = {3, -2, 4}
Output: 3 4
Explanation: The first number will cancel out the second number.
Hence, the output array will be {3,4}.

Input: integers[] = {-2, -1, 1, 2}
Output: -2 -1 1 2
Explanation: After a positive number, negative number is not added.
So every number would be present in the output array

Approach: Ignore numbers of the same sign irrespective of their magnitude. So the only case that we have to consider is when the previous element is of positive magnitude and the next element is of negative magnitude We can use stack to simulate the problem. Follow the steps below to solve the problem:

• Initialize the stack s[].
• Iterate over the range [0, N) using the variable i and perform the following tasks:
• If integers[i] is greater than 0 or s[] is empty, then push integers[i] into the stack s[].
• Otherwise, traverse in a while loop till s is not empty and s.top() is greater than 0 and s.top() is less than abs(integers[i]) and perform the following tasks:
• Pop the element from the stack s[].
• If s is not empty and s.top() equals abs(integers[i]) then pop from the stack s[].
• Else if s[] is empty or s.top() is less than 0 then push integers[i] into the stack s[].
• Initialize the vector res[s.size()] to store the result.
• Iterate over the range [s.size()-1, 0] using the variable i and perform the following tasks:
• Set res[i] as s.top() and pop from the stack s[].
• After performing the above steps, print the value of res[] 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 the remaining numbers` `vector<``int``> remainingNumbers(vector<``int``>& ` `                             ``integers)` `{`   `    ``// Size of the array` `    ``int` `n = integers.size();`   `    ``// Initialize the stack` `    ``stack<``int``> s;`   `    ``// Traverse the array` `    ``for` `(``int` `i = 0; i < n; i++) {` `        ``if` `(integers[i] > 0 || s.empty()) {` `            ``s.push(integers[i]);` `        ``}` `        ``else` `{` `            ``while` `(!s.empty() and s.top() > 0` `                   ``and s.top() < ` `                   ``abs``(integers[i])) {` `                ``s.pop();` `            ``}` `            ``if` `(!s.empty()` `                ``and s.top() == ` `                ``abs``(integers[i])) {` `                ``s.pop();` `            ``}` `            ``else` `if` `(s.empty() || ` `                     ``s.top() < 0) {` `                ``s.push(integers[i]);` `            ``}` `        ``}` `    ``}`   `    ``// Finally we are returning the elements` `    ``// which remains in the stack.` `    ``// we have to return them in reverse order.` `    ``vector<``int``> res(s.size());` `    ``for` `(``int` `i = (``int``)s.size() - 1; i >= 0; ` `         ``i--) {` `        ``res[i] = s.top();` `        ``s.pop();` `    ``}` `    ``return` `res;` `}`   `// Driver Code` `int` `main()` `{`   `    ``vector<``int``> integers = { 3, -2, 4 };` `    ``vector<``int``> ans = ` `        ``remainingNumbers(integers);`   `    ``for` `(``int` `x : ans) {` `        ``cout << x << ``" "``;` `    ``}`   `    ``return` `0;` `}`

## Java

 `// Java program for the above approach`   `import` `java.util.Stack;`   `class` `GFG {`   `    ``// Function to find the remaining numbers` `    ``static` `int``[] remainingNumbers(``int``[] integers)` `    ``{`   `        ``// Size of the array` `        ``int` `n = integers.length;`   `        ``// Initialize the stack` `        ``Stack s = ``new` `Stack();`   `        ``// Traverse the array` `        ``for` `(``int` `i = ``0``; i < n; i++) {` `            ``if` `(integers[i] > ``0` `|| s.empty()) {` `                ``s.push(integers[i]);` `            ``}` `            ``else` `{` `                ``while` `(!s.empty() && s.peek() > ``0` `                       ``&& s.peek()` `                              ``< Math.abs(integers[i])) {` `                    ``s.pop();` `                ``}` `                ``if` `(!s.empty()` `                    ``&& s.peek() == Math.abs(integers[i])) {` `                    ``s.pop();` `                ``}` `                ``else` `if` `(s.empty() || s.peek() < ``0``) {` `                    ``s.push(integers[i]);` `                ``}` `            ``}` `        ``}`   `        ``// Finally we are returning the elements` `        ``// which remains in the stack.` `        ``// we have to return them in reverse order.` `        ``int``[] res = ``new` `int``[s.size()];` `        ``for` `(``int` `i = s.size() - ``1``; i >= ``0``; i--) {` `            ``res[i] = s.peek();` `            ``s.pop();` `        ``}` `        ``return` `res;` `    ``}`   `    ``// Driver Code` `    ``public` `static` `void` `main(String args[])` `    ``{`   `        ``int``[] integers = { ``3``, -``2``, ``4` `};` `        ``int``[] ans = remainingNumbers(integers);`   `        ``for` `(``int` `x : ans) {` `            ``System.out.print(x + ``" "``);` `        ``}` `    ``}` `}`   `// This code is contributed by Lovely Jain`

## Python3

 `# python3 program for the above approach`   `# Function to find the remaining numbers`     `def` `remainingNumbers(integers):`   `        ``# Size of the array` `    ``n ``=` `len``(integers)`   `    ``# Initialize the stack` `    ``s ``=` `[]`   `    ``# Traverse the array` `    ``for` `i ``in` `range``(``0``, n):` `        ``if` `(integers[i] > ``0` `or` `len``(s) ``=``=` `0``):` `            ``s.append(integers[i])`   `        ``else``:` `            ``while` `(``len``(s) !``=` `0` `and` `s[``len``(s) ``-` `1``] > ``0` `                   ``and` `s[``len``(s) ``-` `1``] < ``abs``(integers[i])):` `                ``s.pop()`   `            ``if` `(``len``(s) !``=` `0` `                ``and` `s[``len``(s) ``-` `1``] ``=``=` `                    ``abs``(integers[i])):` `                ``s.pop()`   `            ``elif` `(``len``(s) ``=``=` `0` `or` `                  ``s[``len``(s) ``-` `1``] < ``0``):` `                ``s.append(integers[i])`   `        ``# Finally we are returning the elements` `        ``# which remains in the stack.` `        ``# we have to return them in reverse order.` `    ``res ``=` `[``0` `for` `_ ``in` `range``(``len``(s))]` `    ``for` `i ``in` `range``(``len``(s) ``-` `1``, ``-``1``, ``-``1``):` `        ``res[i] ``=` `s[``len``(s) ``-` `1``]` `        ``s.pop()`   `    ``return` `res`     `# Driver Code` `if` `__name__ ``=``=` `"__main__"``:`   `    ``integers ``=` `[``3``, ``-``2``, ``4``]` `    ``ans ``=` `remainingNumbers(integers)`   `    ``for` `x ``in` `ans:` `        ``print``(x, end``=``" "``)`   `    ``# This code is contributed by rakeshsahni`

## C#

 `// C# program for the above approach` `using` `System;` `using` `System.Collections.Generic;` `class` `GFG` `{` `  `  `  ``// Function to find the remaining numbers` `  ``static` `List<``int``> remainingNumbers(List<``int``> integers)` `  ``{`   `    ``// Size of the array` `    ``int` `n = integers.Count;`   `    ``// Initialize the stack` `    ``Stack<``int``> s = ``new` `Stack<``int``>();`   `    ``// Traverse the array` `    ``for` `(``int` `i = 0; i < n; i++) {` `      ``if` `(integers[i] > 0 || s.Count == 0) {` `        ``s.Push(integers[i]);` `      ``}` `      ``else` `{` `        ``while` `(s.Count != 0 && s.Peek() > 0` `               ``&& s.Peek()` `               ``< Math.Abs(integers[i])) {` `          ``s.Pop();` `        ``}` `        ``if` `(s.Count != 0` `            ``&& s.Peek() == Math.Abs(integers[i])) {` `          ``s.Pop();` `        ``}` `        ``else` `if` `(s.Count == 0 || s.Peek() < 0) {` `          ``s.Push(integers[i]);` `        ``}` `      ``}` `    ``}`     `    ``// Finally we are returning the elements` `    ``// which remains in the stack.` `    ``// we have to return them in reverse order.` `    ``List<``int``> res = ``new` `List<``int``>(``new` `int` `[s.Count]);`   `    ``for` `(``int` `i = (``int``)s.Count - 1; i >= 0; i--) {` `      ``res[i] = s.Peek();` `      ``s.Pop();` `    ``}` `    ``return` `res;` `  ``}`   `  ``// Driver Code` `  ``public` `static` `void` `Main()` `  ``{`   `    ``List<``int``> integers = ``new` `List<``int``>() { 3, -2, 4 };` `    ``List<``int``> ans = remainingNumbers(integers);`   `    ``foreach``(``int` `x ``in` `ans) { Console.Write(x + ``" "``); }` `  ``}` `}`   `// This code is contributed by ukasp.`

## Javascript

 `  ```

Output

`3 4 `

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

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