Akamai Interview Question for Software Engineers


Country: United States
Interview Type: In-Person




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of 0 vote

Use Hashmap while traversing through array, first element already present in hashmap is the duplicate. Sharing my solution in Python below.

def secondDuplicate(nums):
  counterMap = {}
  for num in nums:
    if num not in counterMap:
      counterMap[num] = 0
    else:
      return num
  return -1

Test Code

print(secondDuplicate([1,2,3,4,5,3,7,2])) # Prints 3

- prudent_programmer September 14, 2018 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 vote

Instead of Using Hashmap we can sort the list and count the second number.

func secondNumber(arr []int) int {
	var sec_count int
	var same_num int
	sort.Ints(arr)
	for i := 0; i < len(arr)-1; i++ {
		fmt.Println(arr[i], sec_count, arr[i])
		if arr[i] == arr[i+1] {
			if sec_count == 1 && arr[i] != same_num {
				return arr[i]
			} else {
				sec_count += 1
				same_num = arr[i]
			}
		}
	}
	return -1
}

- testdata017 September 16, 2018 | Flag Reply
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of 0 vote

def secDuplicate(li):
    first = None
    sec = None
    for i, val in enumerate (li):
        if val in li [i+1:]:
            if first == None:
                first = val 
            else:
                sec = val 
                break
    return (first, sec)

testCases = {"no_dups": [1,7,19,8,6,5,9,2, 3], "one_dup": [5,6,3,2,1,1,9], "two_dups": [2,1,3,1,5,2,7]}
for k, li in testCases.items():
    print (k, li, secDuplicate (li))

no_dups [1, 7, 19, 8, 6, 5, 9, 2, 3] (None, None)
one_dup [5, 6, 3, 2, 1, 1, 9] (1, None)
two_dups [2, 1, 3, 1, 5, 2, 7] (2, 1)

- SD September 20, 2018 | Flag Reply


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