Amazon Interview Question for SDE-2s


Country: India
Interview Type: In-Person




Comment hidden because of low score. Click to expand.
7
of 7 vote

Scan from Right for the O(N) Solution.
Scan all the elements from right to left in array and keep track of maximum till now. When maximum changes it’s value, print the same as Leader.

void FindLeader ( int Arr[], int ArrSize ) {
	int max = Arr[ArrSize-1] ;
	for ( int i = ArrSize-1; i >= 0; i– ) {
		if ( Arr[i] >= max ) {
			cout << Arr[i] << ” ” ;
			max = Arr[i] ;
		}
	}
	cout << endl ;
}

Time Complexity: O(N)

- R@M3$H.N October 03, 2014 | Flag Reply
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0
of 0 votes

This is smart. The problem become like a local max problem.

- Anonymous January 27, 2015 | Flag
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7
of 7 vote

private static void findLeader(int[] arr) {
		int max=Integer.MIN_VALUE;
		System.out.println("----Leaders are-----");
		for(int i=arr.length-1;i>=0;i--)
		{
			if(arr[i]>max)
			{
				System.out.print("  "+arr[i]);
				max=arr[i];
			}
		}
	}

- Vir Pratap Uttam May 13, 2015 | Flag Reply
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1
of 1 vote

void findLeaders ( int[] arr) {
	for(int i=0;i<arr.length;i++){
		for(int j=i+1;j<arr.length;j++){
			if(arr[i]>=arr[j]){
				if(j==arr.length-1){
					System.out.print(arr[i]);
					break;
				}
				continue;
			}
			break;
		}
	}
	System.out.print(arr[arr.length-1]);
}

- Neetesh Bhardwaj October 07, 2014 | Flag Reply
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0
of 0 vote

private static void findArrayLeaders(int arr[]) {
		if (arr == null || arr.length == 0) {
			System.out.println("Invalid Inputs");
			return;
		}
		int i = 0, len = 0, leader = 0;
		len = arr.length;
		leader = arr[len - 1];
		System.out.print(leader);
		for (i = len - 2; i >= 0; i--) {
			if (arr[i] > leader) {
				leader = arr[i];
				System.out.print(" " + leader);
			}
		}
	}

- Anonymous October 03, 2014 | Flag Reply
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0
of 0 vote

function findLeader(inputArr) {
    var leaderArray = [];

    for (var i = 0; i < inputArr.length; i++) {
        if (isLeader(inputArr, i)) {
            leaderArray.push(inputArr[i]);
        }
    }
    return leaderArray;
}

function isLeader(inputArr, pos) {
    for (var j = pos; j < inputArr.length; j++) {
        if (inputArr[pos] < inputArr[j]) {
            return false;
        }
    }
     return true;
}

- john madden john madden john madden October 03, 2014 | Flag Reply
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0
of 0 vote

public class GetArrLeader {

    public static void main(String[] args) {
    int[] intArr = {12,14,10,12,2};
    // should print 14,12 and 2
    printLeaders(intArr);
    }

    private static void printLeaders(int[] intArr) {

        for(int i=0;i<(intArr.length)-1;i++) {
             int j = i+i;
             boolean bool = false;
              while (j<intArr.length) {
                  if(intArr[i]>intArr[j++])
                      bool = true;
                  else {
                      bool = false;
                      break;
                  }
              }
            if(bool) System.out.println(intArr[i]);

        }
        System.out.println(intArr[intArr.length-1]); // last one is always considered
    }

}

- Prashant D October 13, 2014 | Flag Reply
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0
of 0 vote

var ele=[13,17,5,4,6,2];
ele=ele.reverse();

var leader_ele=ele[0], leaders=[];
ele.map( function(item) {
if (leader_ele<=item){
leader_ele=item;
leaders.push(leader_ele);
}

})
console.log(leaders.reverse());

- Sathish Kumar S October 29, 2014 | Flag Reply
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0
of 0 vote

Javascript code, you can run in browser console. just copy and paste it in browser console to see output.

var ele=[13,17,5,4,6,2];
ele=ele.reverse();

   var leader_ele=ele[0], leaders=[];
    ele.map( function(item) {
     if (leader_ele<=item){
        leader_ele=item;
        leaders.push(leader_ele);
     }

})
console.log(leaders.reverse());

- Sathish Kumar S October 29, 2014 | Flag Reply
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0
of 0 vote

int main() {
  int arr[6] = {13, 17, 5, 4, 6, 2};
  int max = 0;
  int index = 5;

  for (; index >= 0; index--) {
    if (arr[index] >= max) {
      printf("%d ", arr[index]);
      max = arr[index];
    }
  }

  return 0;

}

- JobSeeker November 07, 2014 | Flag Reply


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