Find middle point segment from given segment lengths

Last Updated : 12 Sep, 2022

Given an array arr[] of size M. The array represents segment lengths of different sizes. These segments divide a line beginning with 0. The value of arr[0] represents a segment from 0 arr[0], value of arr[1] represents segment from arr[0] to arr[1], and so on.
The task is to find the segment which contains the middle point, If the middle segment does not exist, print '-1'.

Examples:

Input: arr = {3, 2, 8}
Output: 3
The three segments are (0, 3), (3, 5), (5, 13)
middle point is 6.5 which is in the 3rd segment.

Input: arr = {3, 2, 5}
Output: -1
Middle point is 5 which is between segments 2 and 3.

Approach: The middle point will always be N / 2. Now, check in which segment does this point exist and print the segment number. If it is the starting or ending for any segment then print '-1'.

Below is the implementation of the above approach:

C++

// C/C++ implementation of the approach
#include <iostream>
using namespace std;

// Function that returns the segment for the
// middle point
int findSegment(int n, int m, int segment_length[])
{

    // the middle point
    double meet_point = (1.0 * n) / 2.0;
    int sum = 0;

    // stores the segment index
    int segment_number = 0;

    for (int i = 0; i < m; i++) {

        // increment sum by
        // length of the segment
        sum += segment_length[i];

        // if the middle is
        // in between two segments
        if ((double)sum == meet_point) {
            segment_number = -1;
            break;
        }

        // if sum is greater
        // than middle point
        if (sum > meet_point) {
            segment_number = i + 1;
            break;
        }
    }

    return segment_number;
}

// Driver code
int main()
{
    int n = 13;
    int m = 3;
    int segment_length[] = { 3, 2, 8 };

    int ans = findSegment(n, m, segment_length);
    cout << (ans);

    return 0;
}

Java

// Java implementation of the approach
class GFG {

    // Function that returns the segment for the
    // middle point
    static int findSegment(int n, int m,
                           int[] segment_length)
    {

        // the middle point
        double meet_point = (1.0 * n) / 2.0;
        int sum = 0;

        // stores the segment index
        int segment_number = 0;

        for (int i = 0; i < m; i++) {

            // increment sum by
            // length of the segment
            sum += segment_length[i];

            // if the middle is
            // in between two segments
            if ((double)sum == meet_point) {
                segment_number = -1;
                break;
            }

            // if sum is greater
            // than middle point
            if (sum > meet_point) {
                segment_number = i + 1;
                break;
            }
        }

        return segment_number;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 13;
        int m = 3;
        int[] segment_length = new int[] { 3, 2, 8 };

        int ans = findSegment(n, m, segment_length);
        System.out.println(ans);
    }
}

Python3

# Python 3 implementation of the approach

# Function that returns the segment for the
# middle point


def findSegment(n, m, segment_length):
    # the middle point
    meet_point = (1.0 * n) / 2.0
    sum = 0

    # stores the segment index
    segment_number = 0

    for i in range(0, m, 1):
        # increment sum by
        # length of the segment
        sum += segment_length[i]

        # if the middle is
        # in between two segments
        if (sum == meet_point):
            segment_number = -1
            break

        # if sum is greater
        # than middle point
        if (sum > meet_point):
            segment_number = i + 1
            break

    return segment_number


# Driver code
if __name__ == '__main__':
    n = 13
    m = 3
    segment_length = [3, 2, 8]

    ans = findSegment(n, m, segment_length)
    print(ans)
# This code is contributed by
# Surendra_Gangwar

// C# implementation of the approach
using System;
class GFG {

    // Function that returns the
    // segment for the middle point
    static int findSegment(int n, int m,
                           int[] segment_length)
    {

        // the middle point
        double meet_point = (1.0 * n) / 2.0;
        int sum = 0;

        // stores the segment index
        int segment_number = 0;

        for (int i = 0; i < m; i++) {

            // increment sum by
            // length of the segment
            sum += segment_length[i];

            // if the middle is
            // in between two segments
            if ((double)sum == meet_point) {
                segment_number = -1;
                break;
            }

            // if sum is greater
            // than middle point
            if (sum > meet_point) {
                segment_number = i + 1;
                break;
            }
        }

        return segment_number;
    }

    // Driver code
    public static void Main()
    {
        int n = 13;
        int m = 3;
        int[] segment_length = new int[] { 3, 2, 8 };

        int ans = findSegment(n, m, segment_length);
        Console.WriteLine(ans);
    }
}

// This code is contributed
// by shs

PHP

<?php
// PHP ementation of the approach 

// Function that returns the segment 
// for the middle point 
function findSegment($n, $m, 
                     $segment_length) 
{ 

    // the middle point 
    $meet_point = (1.0 * $n) / 2.0; 
    $sum = 0; 

    // stores the segment index 
    $segment_number = 0; 

    for ($i = 0; $i < $m; $i++) 
    { 

        // increment sum by 
        // length of the segment 
        $sum += $segment_length[$i]; 

        // if the middle is 
        // in between two segments 
        if ((double)$sum == $meet_point)
        { 
            $segment_number = -1; 
            break; 
        } 

        // if sum is greater 
        // than middle point 
        if ($sum > $meet_point) 
        { 
            $segment_number = $i + 1; 
            break; 
        } 
    } 

    return $segment_number; 
} 

// Driver code 
$n = 13; 
$m = 3; 
$segment_length = array( 3, 2, 8 ); 

$ans = findSegment($n, $m, 
                   $segment_length); 
echo ($ans); 
    
// This code is contributed by ajit
?>

JavaScript

<script>
// Javascript implementation of the approach

    // Function that returns the segment for the
    // middle point
    function findSegment( n, m ,segment_length) {

        // the middle point
        let meet_point = (1.0 * n) / 2.0;
        let sum = 0;

        // stores the segment index
        let segment_number = 0, i;

        for ( i = 0; i < m; i++) {

            // increment sum by
            // length of the segment
            sum += segment_length[i];

            // if the middle is
            // in between two segments
            if ( sum == meet_point) {
                segment_number = -1;
                break;
            }

            // if sum is greater
            // than middle point
            if (sum > meet_point) {
                segment_number = i + 1;
                break;
            }
        }
        return segment_number;
    }

    // Driver code     
    let n = 13;
    let m = 3;
    let segment_length =[ 3, 2, 8 ];

    let ans = findSegment(n, m, segment_length);
    document.write(ans);

// This code is contributed by Rajput-Ji
</script>

Output

Complexity Analysis:

Time Complexity: O(m), for traversal
Auxiliary Space: O(1), as no extra space is required

Analysis of Algorithms

AmanKumarSingh

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Find middle point segment from given segment lengths

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