Bloomberg LP Interview Question for Senior Software Development Engineers


Country: United States
Interview Type: In-Person




Comment hidden because of low score. Click to expand.
2
of 4 vote

a is the array and k is the given value
1- Sort the array
2- Assign two pointer start=0 and end=a.length-1
3- run loop while start < end
a- if(a[start] + a[end] == k), print pair values at start and end index, start++,end--
b- else if(a[start] + a[end] < k) , start++
c- else if(a[start] + a[end] > k) , end--

Time complexity O(nlogn), The above logic will work with all types of data.

- azambhrgn May 09, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
1
of 1 vote

1- In first pass use a hashmap to add all elements to that hashmap.
2- in the second pass. check if the hashmap.Contains(Sum-array[i]), If yes you have the pair.

Time Complexity O(n)

Psuedo code. You can use C# generics or C++ template to make it generic.

var dict = new Dictonary(int,int)
   for(int i = 0; i < array.length; i ++)
{
    if(!dict.Contains(array[i])) 
   {
       dict.Add(array[i],array[i])
   }
}

for(int i = 0 ; i < array.length;  i++)
{
    if(dict.Contains(Sum-arr[i]))
    {
              Console.Writeline( "{" + arr[i] + "," + (Sum-arr[i]).ToString() + "}");
    }
}

~Cheers

- techbits May 14, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

package org.test.combo;

public class TestCombo {
	public static void main(String[] args) {
		int num = 7;

		int[] numArr = { 2, 1, 9, 5, 3, 6 };
		for (int i = 0; i <= numArr.length - 1; i++) {
			for (int j = i + 1; j <= numArr.length - 1; j++) {
				if ((numArr[i] + numArr[j]) == num) {
					System.out.print("[(" + numArr[i] + "," + numArr[j] + ")]");
				}
			}
		}

	}
}

- JamshedK May 18, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

package org.test.combo;

public class TestCombo {
	public static void main(String[] args) {
		int num = 7;

		int[] numArr = { 2, 1, 9, 5, 3, 6 };
		for (int i = 0; i <= numArr.length - 1; i++) {
			for (int j = i + 1; j <= numArr.length - 1; j++) {
				if ((numArr[i] + numArr[j]) == num) {
					System.out.print("[(" + numArr[i] + "," + numArr[j] + ")]");
				}
			}
		}

	}
}

- JamshedK May 18, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

if repetition is allowed in the input data set, bellow

template <typename T>
std::vector<std::pair<T, T>> printEqualPairs2(std::vector<T> vec, int sum)
{
std::vector<std::pair<T, T>> ret;
std::multiset<T> mulSet;
for (auto i : vec)
mulSet.emplace(i);

for (auto i : vec)
{
auto it = mulSet.find(sum - i);
if (it != mulSet.end())
{
std::pair<T, T> pair(i, *it);
ret.push_back(pair);
mulSet.erase(it);
auto it2 = mulSet.find(i);
if (it2 != mulSet.end())
mulSet.erase(it2);
}
}
return ret;
}

- janitha June 11, 2017 | Flag Reply
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0
of 0 vote

template <typename T>
std::vector<std::pair<T,T>> findPairs(const std::vector<T>& first, T k)
{
    std::unordered_set<T> firstSet(first.begin(), first.end());
    std::vector<std::pair<T,T>> result;
    
    for(const auto& val: first)
    {
        auto found = firstSet.find(k-val);
        if( found != firstSet.end())
        {
            result.emplace_back(std::pair<T,T>(*found, val));
            firstSet.erase(*found);
            firstSet.erase(val);
        }
        
    }
    return result;
    
}

int main()
{
    
    std::vector<int> first={1,2,3,4,5,7};
    std::vector<std::pair<int,int>> result = findPairs( first, 3);
    
    for( const auto& ele: result)
    {
        std::cout << " first: " << ele.first << "second: "<< ele.second << "\n";
    }

}

- Anonymous June 28, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void findSumHashMap(int[] arr, int k){
		
	    Map<Integer, Integer> pairs = new HashMap<Integer, Integer>();

	    for(int i=0;i<arr.length;i++){
	    	
	    	if(pairs.containsKey(arr[i])){
	    		
	    		System.out.println(arr[i]+", "+pairs.get(arr[i]));
	    	}else{
	    		
	    		pairs.put(k-arr[i], arr[i]);
	    	}
	    }
	}

- Anonymous July 29, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

public static void findSumHashMap(int[] arr, int k){
		
	    Map<Integer, Integer> pairs = new HashMap<Integer, Integer>();

	    for(int i=0;i<arr.length;i++){
	    	
	    	if(pairs.containsKey(arr[i])){
	    		
	    		System.out.println(arr[i]+", "+pairs.get(arr[i]));
	    	}else{
	    		
	    		pairs.put(k-arr[i], arr[i]);
	    	}
	    }
	}

- ag91375 July 29, 2017 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

In python.

Correct me if am wrong

import json
x1=input("Enter the list:")
x=json.loads(x1)
y=int(input("enter the sum value:"))
setted=set()
#sett=set()
def sum_array(list1,value):
    x=0
    for i in range(0,len(list1)):
        for j in range(0,len(list1)):
            if i!=j:
                x=list1[i]+list1[j]
                if x==value:
                    setted.add((list1[i],list1[j]))
                   
    print("The repeated item is pair is ",setted)
sum_array(x,y)

- jagannathans92 February 20, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

<T> void pairs(final T[] nums, T sum) {
	if (nums == null) return;
	final Set<T> numsLeft = Arrays.stream(nums).boxed().collect(Collectors.toSet());
	
	for (int i = 0; i < nums.length; ++i) {
		final T diff = sum - nums[i];
		if (numsLeft.contains(diff)) {
			System.out.println(String.format("%d, %d", nums[i], numsLeft.get(diff)));
			numsLeft.remove(nums[i]);
			numsLeft.remove(diff);
		}
	}
}

- jtcgen April 05, 2018 | Flag Reply


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