Find element at given index after given range reversals

An array consisting of N elements is given. There are several reversals we do in unique ranges[L..R]. The task is to print the element at given index.

Examples:

Input : 
arr[] : 10 20 30 40 50
ranges[] = {{1, 4}, {0, 2}}
Query Index = 1
Output : 50
Explanation : 
Reverse range[1..4] : 10 50 40 30 20
Reverse range[0..2] : 40 50 10 30 20
So we have 50 at index 1

The Brute Force approach would be to actually reverse a range of elements and answer the queries afterward.

Efficient Method: If we observe, the reversal of range[L..R] will result as follows : 
The index will become right + left – index. 
By doing this, we can compute the index easily. 

Implementation:

C++

// Program to find index of an element after
// given range reversals.
#include <bits/stdc++.h>
using namespace std;

// Function to compute the element at query index
int answer(int arr[], int ranges[][2], int reversals,
           int index)
{
    for (int i = reversals - 1; i >= 0; i--) {
        // Range[left...right]
        int left = ranges[i][0], right = ranges[i][1];

        // If doesn't satisfy, reversal won't
        // have any effect
        if (left <= index && right >= index)
            index = right + left - index;
    }

    // Returning element at modified index
    return arr[index];
}

// Driver
int main()
{
    int arr[] = { 10, 20, 30, 40, 50 };

    // reversals
    int reversals = 2;
    int ranges[reversals][2] = { { 1, 4 }, { 0, 2 } };

    int index = 1;
    cout << answer(arr, ranges, reversals, index);

    return 0;
}

Java

// Program to find index of an element
// after given range reversals.
import java.util.Arrays;

class GFG {
    // Function to compute the element at
    // query index
    static int answer(int[] arr, int[][] ranges,
                      int reversals, int index)
    {
        for (int i = reversals - 1; i >= 0; i--) {
            // Range[left...right]
            int left = ranges[i][0],
                right = ranges[i][1];

            // If doesn't satisfy, reversal
            // won't have any effect
            if (left <= index && right >= index)
                index = right + left - index;
        }

        // Returning element at modified index
        return arr[index];
    }

    // Driver code
    public static void main(String[] args)
    {

        int[] arr = { 10, 20, 30, 40, 50 };

        // reversals
        int reversals = 2;
        int[][] ranges = { { 1, 4 }, { 0, 2 } };

        int index = 1;
        System.out.println(answer(arr, ranges,
                                  reversals, index));
    }
}

/* This code is contributed by Mr. Somesh Awasthi */

Python3

# Program to find index of an element
# after given range reversals.

# Function to compute the element
# at query index
def answer(arr, ranges, reversals, index):
    i = reversals - 1
    while(i >= 0):

        # Range[left...right]
        left = ranges[i][0]
        right = ranges[i][1]

        # If doesn't satisfy, reversal won't
        # have any effect
        if (left <= index and right >= index):
            index = right + left - index

        i -= 1

    # Returning element at modified index
    return arr[index]

# Driver Code
if __name__ == '__main__':
    arr = [10, 20, 30, 40, 50]

    # reversals
    reversals = 2
    ranges = [ [ 1, 4 ], [ 0, 2 ] ]

    index = 1
    print(answer(arr, ranges,
                 reversals, index))

# This code is contributed by
# Surendra_Gangwar

C#

// C# program to find index of an element
// after given range reversals.
using System;

class GFG {

    // Function to compute the element at
    // query index
    static int answer(int[] arr, int[, ] ranges,
                       int reversals, int index)
    {
        for (int i = reversals - 1; i >= 0; i--)
        {

            // Range[left...right]
            int left = ranges[i, 0],
                right = ranges[i, 1];

            // If doesn't satisfy, reversal
            // won't have any effect
            if (left <= index && right >= index)
                index = right + left - index;
        }

        // Returning element at modified index
        return arr[index];
    }

    // Driver code
    public static void Main()
    {

        int[] arr = { 10, 20, 30, 40, 50 };

        // reversals
        int reversals = 2;
        int[, ] ranges = { { 1, 4 }, { 0, 2 } };

        int index = 1;
        Console.WriteLine(answer(arr, ranges,
                                reversals, index));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// Program to find index
// of an element after
// given range reversals.

// Function to compute the
// element at query index
function answer($arr, $ranges,
                $reversals, $index)
{
    for ($i = $reversals - 1;
              $i >= 0; $i--)
    {
        // Range[left...right]
        $left = $ranges[$i][0];
        $right = $ranges[$i][1];

        // If doesn't satisfy,
        // reversal won't have
        // any effect
        if ($left <= $index &&
            $right >= $index)
            $index = $right + $left -
                              $index;
    }

    // Returning element
    // at modified index
    return $arr[$index];
}

// Driver Code
$arr = array( 10, 20, 30, 40, 50 );

// reversals
$reversals = 2;
$ranges = array(array( 1, 4 ),
                array( 0, 2 ));

$index = 1;
echo answer($arr, $ranges,
            $reversals, $index);

// This code is contributed
// by nitin mittal.
?>

Javascript

<script>

    // Program to find index of an element
    // after given range reversals.

    // Function to compute the element at
    // query index
    function answer(arr, ranges, reversals, index)
    {
        for (let i = reversals - 1; i >= 0; i--) {
            // Range[left...right]
            let left = ranges[i][0],
                right = ranges[i][1];

            // If doesn't satisfy, reversal
            // won't have any effect
            if (left <= index && right >= index)
                index = right + left - index;
        }

        // Returning element at modified index
        return arr[index];
    }

    let arr = [ 10, 20, 30, 40, 50 ];

    // reversals
    let reversals = 2;
    let ranges = [ [ 1, 4 ], [ 0, 2 ] ];

    let index = 1;
    document.write(answer(arr, ranges, reversals, index));

</script>

Output

50

Optimized solution

We can start by applying all the reversals to the array in the order they are given. To do this efficiently, we can use a helper function to reverse a given range of elements in the array. After applying all the reversals, we can answer the queries directly by accessing the element at the given index.

Algorithm

First define function reverseRange(arr, L, R) that takes an array arr such that
   a.Two indices L and R as input.
If L < R, THAN swap the elements at indices L and R in the array arr.
After that Increment L and decrement R.
Define a function applyReversals(arr, ranges) 
For each pair of indices (L, R) in ranges, call the reverseRange function with arguments (arr, L, R).
Define a function getElementAtIndex(arr, index)  
Return the element of arr at index index.
Initialize an array arr and a list of ranges ranges.
Call the applyReversals and getElementAtIndex function 
Assign the result to variable result.
Get result as output.

Implementation of above program

C++

#include <iostream>
#include <vector>
using namespace std;

void reverseRange(vector<int>& arr, int L, int R) {
    while (L < R) {
        swap(arr[L], arr[R]);
        L++;
        R--;
    }
}

void applyReversals(vector<int>& arr, vector<pair<int, int>>& ranges) {
    for (int i = 0; i < ranges.size(); i++) {
        int L = ranges[i].first;
        int R = ranges[i].second;
        reverseRange(arr, L, R);
    }
}

int getElementAtIndex(vector<int>& arr, int index) {
    return arr[index];
}

int main() {
    // initialize inputs
    vector<int> arr = {10, 20, 30, 40, 50};
    vector<pair<int, int>> ranges = {{1, 4}, {0, 2}};
    int queryIndex = 1;

    // apply reversals and answer query
    applyReversals(arr, ranges);
    int result = getElementAtIndex(arr, queryIndex);

    // output result
    cout << result << endl;

    return 0;
}

Python

def reverseRange(arr, L, R):
    while L < R:
        arr[L], arr[R] = arr[R], arr[L]
        L += 1
        R -= 1

def applyReversals(arr, ranges):
    for L, R in ranges:
        reverseRange(arr, L, R)

def getElementAtIndex(arr, index):
    return arr[index]

# initialize inputs
arr = [10, 20, 30, 40, 50]
ranges = [(1, 4), (0, 2)]
queryIndex = 1

# apply reversals and answer query
applyReversals(arr, ranges)
result = getElementAtIndex(arr, queryIndex)

# output result
print(result)

Output

50

 Time complexity  O(N*M), where N is the length of the input array arr and M is the number of ranges in the input ranges. 

 Space complexity is O(1),as it does not use any additional memory. 

This article is contributed by Rohit Thapliyal. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.