Non-Overlapping intervals program in Go (Golang)

Table of Contents

Overview

An array of intervals is given where intervals[i] = [starti, endi]. We have to find out the minimum number of intervals to remove so that the interval in the intervals array become non-overlapping

Let’s understand it with an example

Input: intervals = [[2,3],[3,4],[4,5],[2,4]]
Output: 1
Explanation: [2,4] can be removed and the rest of the intervals are non-overlapping.

The idea is to first sort based interval start time and then count the overlapping intervals.

Program

Here is the program for the same.

package main
import (
"fmt"
"sort"
)
func eraseOverlapIntervals(intervals [][]int) int {
lenIntervals := len(intervals)
sort.Slice(intervals, func(i, j int) bool {
return intervals[i][0] < intervals[j][0]
})
prevIntervalEnd := intervals[0][1]
minIntervals := 0
for i := 1; i < lenIntervals; i++ {
currentIntervalStart := intervals[i][0]
currentIntervalEnd := intervals[i][1]
if currentIntervalStart < prevIntervalEnd {
minIntervals++
if prevIntervalEnd >= currentIntervalEnd {
prevIntervalEnd = currentIntervalEnd
}
} else {
prevIntervalEnd = currentIntervalEnd
}
}
return minIntervals
}
func main() {
output := eraseOverlapIntervals([][]int{{2, 3}, {3, 4}, {4, 5}, {2, 4}})
fmt.Println(output)
}

Output

6
13

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Also if you are interested in understanding how all design patterns can be implemented in Golang. If yes, then this post is for you –All Design Patterns Golang

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