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

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

Arnab Chakraborty

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

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