Find middle point segment from given segment lengths
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++
#include <iostream>
using namespace std;
int findSegment( int n, int m, int segment_length[])
{
double meet_point = (1.0 * n) / 2.0;
int sum = 0;
int segment_number = 0;
for ( int i = 0; i < m; i++) {
sum += segment_length[i];
if (( double )sum == meet_point) {
segment_number = -1;
break ;
}
if (sum > meet_point) {
segment_number = i + 1;
break ;
}
}
return segment_number;
}
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
class GFG {
static int findSegment( int n, int m,
int [] segment_length)
{
double meet_point = ( 1.0 * n) / 2.0 ;
int sum = 0 ;
int segment_number = 0 ;
for ( int i = 0 ; i < m; i++) {
sum += segment_length[i];
if (( double )sum == meet_point) {
segment_number = - 1 ;
break ;
}
if (sum > meet_point) {
segment_number = i + 1 ;
break ;
}
}
return segment_number;
}
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
def findSegment(n, m, segment_length):
meet_point = ( 1.0 * n) / 2.0
sum = 0
segment_number = 0
for i in range ( 0 , m, 1 ):
sum + = segment_length[i]
if ( sum = = meet_point):
segment_number = - 1
break
if ( sum > meet_point):
segment_number = i + 1
break
return segment_number
if __name__ = = '__main__' :
n = 13
m = 3
segment_length = [ 3 , 2 , 8 ]
ans = findSegment(n, m, segment_length)
print (ans)
|
C#
using System;
class GFG {
static int findSegment( int n, int m,
int [] segment_length)
{
double meet_point = (1.0 * n) / 2.0;
int sum = 0;
int segment_number = 0;
for ( int i = 0; i < m; i++) {
sum += segment_length[i];
if (( double )sum == meet_point) {
segment_number = -1;
break ;
}
if (sum > meet_point) {
segment_number = i + 1;
break ;
}
}
return segment_number;
}
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);
}
}
|
PHP
<?php
function findSegment( $n , $m ,
$segment_length )
{
$meet_point = (1.0 * $n ) / 2.0;
$sum = 0;
$segment_number = 0;
for ( $i = 0; $i < $m ; $i ++)
{
$sum += $segment_length [ $i ];
if ((double) $sum == $meet_point )
{
$segment_number = -1;
break ;
}
if ( $sum > $meet_point )
{
$segment_number = $i + 1;
break ;
}
}
return $segment_number ;
}
$n = 13;
$m = 3;
$segment_length = array ( 3, 2, 8 );
$ans = findSegment( $n , $m ,
$segment_length );
echo ( $ans );
?>
|
Javascript
<script>
function findSegment( n, m ,segment_length) {
let meet_point = (1.0 * n) / 2.0;
let sum = 0;
let segment_number = 0, i;
for ( i = 0; i < m; i++) {
sum += segment_length[i];
if ( sum == meet_point) {
segment_number = -1;
break ;
}
if (sum > meet_point) {
segment_number = i + 1;
break ;
}
}
return segment_number;
}
let n = 13;
let m = 3;
let segment_length =[ 3, 2, 8 ];
let ans = findSegment(n, m, segment_length);
document.write(ans);
</script>
|
Complexity Analysis:
- Time Complexity: O(m), for traversal
- Auxiliary Space: O(1), as no extra space is required
Last Updated :
12 Sep, 2022
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