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  1. Prepare
  2. Algorithms
  3. Greedy
  4. Maximum Perimeter Triangle
  5. Discussions

Maximum Perimeter Triangle

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324 Discussions

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  • gschoen
     + 0 comments

    This question can easily be solved by sorting the sticks array but when you are likely to only examine the top few elements in the array, I prefer to use a priority queue for faster run time.

    void maxPerimeter(int *sticks, int n, int *ans) {
    heapify(sticks, n);
    ans[2] = sticks[0];
    popTop(sticks, &n);
    ans[1] = sticks[0];
    popTop(sticks, &n);
    ans[0] = sticks[0];
    popTop(sticks, &n);
    while (ans[0] + ans[1] <= ans[2]) {
        if (n == 0) {
            ans[0] = -1;
            return;
        }
        ans[2] = ans[1];
        ans[1] = ans[0];
        ans[0] = sticks[0];
        popTop(sticks, &n);
    }
}

  • vutrantrung14821
     + 0 comments
    def maximumPerimeterTriangle(sticks):
    sticks.sort()
    b=sticks[0]+sticks[1]
    a=[0,0,0]
    for i in range(2,n):
        if b>sticks[i]:
            a=[sticks[i-2],sticks[i-1],sticks[i]]
        b=b-sticks[i-2]+sticks[i]
    if a[0]==a[1]==a[2]==0:
        return ['-1']
    else:
        return a

  • alban_tyrex
     + 0 comments

    Here is my c++ solution, you can watch the explanation here : https://youtu.be/cP4PLTgC1OE

    vector<int> maximumPerimeterTriangle(vector<int> sticks) {
    sort(sticks.begin(), sticks.end());
    for(int i = sticks.size() - 1; i >= 2; i--){
        if(sticks[i] < sticks[i - 1] + sticks[i - 2]) return {sticks[i - 2], sticks[i -1], sticks[i] };
    }
    return {-1};
}

  • emhhacker
     + 0 comments

    My Java solution with o(n log n) time complexity and o(1) space complexity:

    public static List<Integer> maximumPerimeterTriangle(List<Integer> sticks) {
        // find 3 vals that produce the max sum; vals are sorted, and first two sum is greater than 3rd val
        
        //sort array ascending
        Collections.sort(sticks);
        
        //check each max len and if it follows the triangle inequality theorem
        for(int i = sticks.size() - 1; i > 1; i--){
            int maxLen = sticks.get(i);
            int midLen = sticks.get(i - 1);
            int minLen = sticks.get(i - 2);
            if(minLen + midLen > maxLen) return Arrays.asList(minLen, midLen, maxLen);
        }
        return Arrays.asList(-1);
    }

  • patrickgao2020
     + 0 comments

    Here is my Python solution!

    def maximumPerimeterTriangle(sticks):
    triangles = []
    for i in range(len(sticks)):
        for j in range(i + 1, len(sticks)):
            for k in range(j + 1, len(sticks)):
                triangle = [sticks[i] ,sticks[j], sticks[k]]
                triangle.sort()
                if sum(triangle[:2]) > triangle[2]:
                    triangles.append([sum(triangle), triangle])
    if triangles:
        return sorted(triangles)[-1][1]
    return ["-1"]