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 pointint 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 codeint 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 approachclass 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 pointdef 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 codeif __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#
// C# implementation of the approachusing 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
3
Complexity Analysis:
- Time Complexity: O(m), for traversal
- Auxiliary Space: O(1), as no extra space is required
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