Day 28 : Fake Coin Problem

Easy

0/40

Contributed by

3 upvotes

Problem statement

You are given an array ‘sum’ which is the prefix sum of an array of coins ‘C’ where ‘C[i]’ is ‘1’ if the coin is real, or ‘0’ if the coin is fake. There is exactly one fake coin in the array.

Return the index of the fake coin. Assume the array to be 0-indexed.

For Example:

‘sum’ = {1,1,2,3} 
Index of the fake coin is 1. 
For the given ‘sum’, ‘C’ will be {1, 0, 1, 1}. Thus the index of the fake coin is 1.

Detailed explanation ( Input/output format, Notes, Images )

Input format:

The first line contains an integer ‘T’, denoting the number of test cases.
For each test case:
The first line contains an integer ‘N’, representing the size of the array ‘sum’.
The second line contains ‘N’ integers, denoting the elements of the array 'sum'.

Output Format:

For each testcase, return the index of the fake coin.

Constraints:

1 <=  ‘T’ <= 10^5
2 <= ‘N’ <= 10^5
0 <= ‘C[i]’ <= 1
0 <= ‘sum[i]’ <= ‘N-1’
Sum of ‘N’ for all testcases <= 10^5
There is exactly one ‘0’ in ‘C’.

Time Limit: 1 sec

Sample Input 1:

Sample Output 1:

2
4

Explanation of sample output 1:

For Test 1:
‘sum’ = {1, 2, 2, 3}
‘C’ = {1, 1, 0, 1}
Index of the fake coin is 2.


For Test 2:
‘sum' = {1, 2, 3, 4, 4}
‘C’ = {1, 1, 1, 1, 0}
Index of the fake coin is 4.

Sample Input 2:

Sample Output 2:

2
1
0

Hint 1

Hint 2

What will be the difference between adjacent elements in sum, if there’s a fake coin present?

Approaches (2)

Approach 1

Approach 2

Linear Search Approach

Approach:

We can simply iterate through all the elements and check for the difference between the current index and the previous index. If there’s a fake coin the difference will be 0. If the first element of ‘sum’ is 0 then the fake coin is at index 0.

Algorithm:

for ‘i’ from 1 to ‘N-1’:
- if(sum[i] == sum[i-1])
  - return i
return 0

Time Complexity

O(N), where ‘N’ is the number of elements in the array ‘sum’.

We are iterating via ‘i’ from 0 to ‘N-1’, hence the overall time complexity of this solution is O(N).

Space Complexity

O(1)

We are not utilising any extra space. Thus the space complexity is constant.

(100% EXP penalty)

Day 28 : Fake Coin Problem

Console