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๐Ÿ”ฐAuthentic Coding Solutions(with Outputs)
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๐—–๐—ฆ ๐—”๐—น๐—ด๐—ผ ๐Ÿ’ป ๐ŸŒ ใ€Ž๐—–๐—ผ๐—บ๐—ฝ๐—ฒ๐˜๐—ถ๐˜๐—ถ๐˜ƒ๐—ฒ ๐—ฃ๐—ฟ๐—ผ๐—ด๐—ฟ๐—ฎ๐—บ๐—บ๐—ถ๐—ป๐—ดใ€
long findSum(long cum[], int l, int r, int n) {     if (l == 0)     {         return cum[r];     }     return cum[r] - cum[l - 1]; } long getThreeNonOverLappingIntervals(vector<int> starting, vector<int> ending) {     vector<pair<int, int>> vp;     int nโ€ฆ
long getThreeNonOverlappingIntervals(vector<int> starting, vector<int> ending)
{
    int n = starting.size();
    vector<int> ends;
    vector<int> starts;
    for (int i = 0; i < n; i++)
    {
        ends.push_back(ending[i]);
        starts.push_back(starting[i]);
    }
    sort(starts.begin(), starts.end());
    sort(ends.begin(), ends.end());
    long ans = 0;
    for (int i = 0; i < n; i++)
    {
        long count = upper_bound(ends.begin(), ends.end(), starting[i] - 1) - ends.begin();
        long count2 = n - (upper_bound(starts.begin(), starts.end(), ending[i]) - starts.begin());
        ans += count * count2;
    }
    return ans;
}

Non Overlapping Intervals โœ…
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def lastElement(A):
    while len(A) > 1:
        B = [A[i] + A[i + 1] for i in range(0, len(A), 2)]
        A = B
    return A[0]

// Last remaining element
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