# Minimum Size of Two Non-Overlapping Intervals in C++

C++Server Side ProgrammingProgramming

Suppose we have a list of intervals where each interval contains the [start, end] times. We have to find the minimum total size of any two non-overlapping intervals, where the size of an interval is (end - start + 1). If we cannot find such two intervals, return 0.

So, if the input is like [[2,5],[9,10],[4,6]], then the output will be 5 as we can pick interval [4,6] of size 3 and [9,10] of size 2.

To solve this, we will follow these steps −

• ret := inf

• n := size of v

• sort the array v based on the end time

• Define an array dp of size n

• for initialize i := 0, when i < size of v, update (increase i by 1), do −

• low := 0, high := i - 1

• temp := inf

• val := v[i, 1] - v[i, 0] + 1

• while low <= high, do

• mid := low + (high - low) / 2

• if v[mid, 1] >= v[i, 0], then −

• high := mid - 1

• Otherwise

• temp := minimum of temp and dp[mid]

• low := mid + 1

• if temp is not equal to inf, then −

• ret := minimum of ret and (temp + val)

• dp[i] := minimum of val and temp

• Otherwise

• dp[i] := val

• if i > 0, then

• dp[i] := minimum of dp[i] and dp[i - 1]

• return (if ret is same as inf, then 0, otherwise ret)

Let us see the following implementation to get better understanding −

## Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
class Solution {
public:
static bool cmp(vector <int>& a, vector <int>& b){
return a[1] < b[1];
}
int solve(vector<vector<int>>& v) {
int ret = INT_MAX;
int n = v.size();
sort(v.begin(), v.end(), cmp);
vector <int> dp(n);
for(int i = 0; i < v.size(); i++){
int low = 0;
int high = i - 1;
int temp = INT_MAX;
int val = v[i][1] - v[i][0] + 1;
while(low <= high){
int mid = low + (high - low) / 2;
if(v[mid][1] >= v[i][0]){
high = mid - 1;
}else{
temp = min(temp, dp[mid]);
low = mid + 1;
}
}
if(temp != INT_MAX){
ret = min(ret, temp + val);
dp[i] = min(val, temp);
}else{
dp[i] = val;
}
if(i > 0) dp[i] = min(dp[i], dp[i - 1]);
}
return ret == INT_MAX ? 0 : ret;
}
};
main(){
Solution ob;
vector<vector<int>> v = {{2,5},{9,10},{4,6}};
cout << (ob.solve(v));
}

## Input

{{2,5},{9,10},{4,6}}

## Output

5
Published on 02-Sep-2020 12:23:03