Non-Overlapping Intervals Problem

Description

LeetCode Problem 435.

Given an array of intervals intervals where intervals[i] = [start_i, end_i], return the minimum number of intervals you need to remove to make the rest of the intervals non-overlapping.

Example 1:

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

Example 2:

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Input: intervals = [[1,2],[1,2],[1,2]]
Output: 2
Explanation: You need to remove two [1,2] to make the rest of the intervals non-overlapping.
``````

Example 3:

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Input: intervals = [[1,2],[2,3]]
Output: 0
Explanation: You don't need to remove any of the intervals since they're already non-overlapping.
``````

Constraints:

• 1 <= intervals.length <= 10^5
• intervals[i].length == 2
• -5 * 10^4 <= start_i < end_i <= 5 * 10^4

Sample C++ Code

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class Solution {
public:
int eraseOverlapIntervals(vector<vector<int>>& intervals) {
int n = intervals.size();
if (n <= 1) return 0;

sort(intervals.begin(), intervals.end());
vector<int> dp(n, 0);

dp[0] = 1;
int ans = 1;
for (int i = 1; i < n; i ++) {
int maxdp = 0;
for (int j = i - 1; j >= 0; j --) {
if (intervals[j][1] <= intervals[i][0]) {
maxdp = max(maxdp, dp[j]);
}
}
dp[i] = maxdp + 1;
ans = max(ans, dp[i]);
}
return n - ans;
}
};
``````