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Program to find removed term from arithmetic sequence in Python
Suppose we have an array called nums holding n-1 arithmetic sequence terms. One element except the first or last element of nums was removed before. We have to find the removed number.
So, if the input is like nums = [5, 7, 11, 13], then the output will be 9 because, the items are following the formula 2i+5, so for i = 2 it will be 2*2 + 5 = 9 which is missing.
To solve this, we will follow these steps −
if size of nums is same as 2, then
return floor of (sum of all elements present in nums)/2
if nums[0] is same as nums[1], then/p>
return nums[0]
lower := nums[0]
upper := last element of nums
interval := floor of (upper - lower) / size of nums
pointer := floor of size of nums / 2
left := 0
right := size of nums - 1
while left is not same as right, do
if nums[pointer] is not same as nums[0] + interval * pointer, then
if nums[pointer - 1] is same as nums[0] + interval *(pointer - 1) , then
return nums[0] + interval * pointer
otherwise,
right := pointer
pointer :=floor of(left + right) / 2
otherwise,
if right - left is same as 1, then
pointer := right
otherwise,
left := pointer
pointer := floor of(left + right) / 2
Example
Let us see the following implementation to get better understanding
def solve(nums):
if len(nums) == 2:
return sum(nums) // 2
if nums[0] == nums[1]:
return nums[0]
lower = nums[0]
upper = nums[-1]
interval = (upper - lower) // len(nums)
pointer = len(nums) // 2
left = 0
right = len(nums) - 1
while left != right:
if nums[pointer] != nums[0] + interval * pointer:
if nums[pointer - 1] == nums[0] + interval * (pointer -1):
return nums[0] + interval * pointer
else:
right = pointer
pointer = (left + right) // 2
else:
if right - left == 1:
pointer = right
else:
left = pointer
pointer = (left + right) // 2
nums = [5, 7, 11, 13]
print(solve(nums))Input
[5, 7, 11, 13]
Output
9
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