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

Maximum Perimeter Triangle

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320 Discussions

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  • 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};
}

  • sreekishore55
     + 0 comments

    Passed 13/13 testcase

    def check1(e):
    return e[3]
def check2(p,d,flag,count):
    for i in p:
        if d==i[3]:
            count+=1
    if count==1:
        return True
    elif count>1:
        return False
p=[]
n=int(input())
data=input().split(" ")
data=[int(i) for i in data]
for i in range(n-2):
    j,k=i+1,i+2
    if data[i]+data[j]>data[k] and data[i]+data[k]>data[j] and data[j]+data[k]>data[i]:
        p.append([data[i],data[j],data[k],data[i]+data[j]+data[k]])
if len(p)<=1:
    if len(p)==0:
        print(-1)
    else:
        print(f'{p[0][0]} {p[0][1]} {p[0][2]}')
else:
    p.sort(reverse=True,key=check1)
    flag=check2(p,p[0][3],True,0)
    if flag==True:
        print(f'{p[0][0]} {p[0][1]} {p[0][2]}')
    else:
        count=0
        flag=check2(p,p[-1][3],True,0)
        if flag==True or flag==False:
            print(f'{p[-1][0]} {p[-1][1]} {p[-1][2]}')

  • pub_ashishk28
     + 0 comments

    python3

    def triangulate(x):
    
    a,b,c = x
    if a + b > c:
        if a + c > b:
            if b + c > a:
                return True
    
    return False

def maximumPerimeterTriangle(sticks):
    # Write your code here
    
    sides = list(combinations(sticks, 3))
    sides = sorted(sides, key=sum, reverse=True)
    sides = list(filter(triangulate, sides))
    
    if not sides:
        return [-1]
        
    valid = list({x for x in sides if sum(x) == sum(sides[0])})
    
    if len(valid) == 1:
        return sorted(valid[0])
        
    valid = sorted(valid, key=max, reverse=True)
    valid =list(filter(lambda x: max(x) == max(valid[0]), valid))
    
    if len(valid) == 1:
        return sorted(valid[0])
        
    valid = sorted(valid, key=min, reverse=True)
    valid = list(filter(lambda x: min(x) == min(valid[0]), valid))
    
    return sorted(valid[0])

  • dorirgob
     + 0 comments

    Java:

    public static List<Integer> maximumPerimeterTriangle(List<Integer> sticks) {
  Collections.sort(sticks);
  int n = sticks.size();
  for (int i = n - 1; i >= 2; i--) {
    int a = sticks.get(i - 2);
    int b = sticks.get(i - 1);
    int c = sticks.get(i);
    if (c < a + b) {
      return Arrays.asList(a, b, c);
    }
  }
  return Arrays.asList(-1);
}

  • ishwardharme320
     + 0 comments

    Java

    public static List maximumPerimeterTriangle(List sticks) {

    // The sum of the lengths of any two sides of a triangle must be greater than the 
//length of the third side.
 int[] arr=new int[3];
 List<Integer> list=new ArrayList<>();
         Collections.sort(sticks);
         for(int i=sticks.size()-1;i>=2;i--){
            arr[0]=sticks.get(i);
            arr[1]=sticks.get(i-1);
            arr[2]=sticks.get(i-2);

            if(arr[0]+arr[1]>arr[2] && arr[1]+arr[2]>arr[0] && arr[2]+arr[0]>arr[1] ){
                list.add(arr[0]);
                list.add(arr[1]);
                list.add(arr[2]);
                break;
            }else{
                continue;
            }
         }

         if(list.size()==0){
            list.add(-1);
           return list;
         }else{
                 Collections.sort(list); 
                 return list;
         }

}