Program to find winner of array removal game in Python

Suppose Amal and Bimal are playing a game where they have one array A with some numbers. The game rules are as follows:

• Bimal will start always
• In each turn one player deletes the maximum element from the array and all other elements present at right of the deleted element will also be deleted
• They play alternatively
• The player who removes all remaining elements wins the game

So, if the input is like nums = [5,2,6,3,4], then the output will be Amal because at first Bimal will remove [6,3,4] so array will be [5,2], then Amal will remove all, so he will be the winner.

Game Strategy

The key insight is that players will always choose the leftmost maximum element from the remaining array. Each time a new maximum is encountered while scanning left to right, it represents a potential move. The number of such moves determines the winner.

Algorithm Steps

To solve this, we will follow these steps −

• maximum := -1
• count := 0
• for each element a in nums, do
  • if a > maximum, then
    • count := count + 1
    • maximum := a
• if count mod 2 is same as 0, then
  • return "Amal"
• return "Bimal"

Example

Let us see the following implementation to get better understanding −

def solve(nums):
    maximum = -1
    count = 0
    for a in nums:
        if a > maximum:
            count += 1
            maximum = a
    
    if count % 2 == 0:
        return "Amal"
    return "Bimal"

nums = [5, 2, 6, 3, 4]
print(f"Input: {nums}")
print(f"Winner: {solve(nums)}")

Input: [5, 2, 6, 3, 4]
Winner: Amal

How It Works

For the array [5,2,6,3,4], we scan left to right:

• 5 > -1, so count = 1 (Bimal's turn)
• 2 < 5, no increment
• 6 > 5, so count = 2 (Amal's turn)
• 3 < 6, no increment
• 4 < 6, no increment

Total moves = 2 (even), so Amal wins.

Additional Example

def solve(nums):
    maximum = -1
    count = 0
    for a in nums:
        if a > maximum:
            count += 1
            maximum = a
    
    if count % 2 == 0:
        return "Amal"
    return "Bimal"

# Test with different arrays
test_cases = [
    [1, 2, 3, 4, 5],
    [5, 4, 3, 2, 1], 
    [3, 1, 4, 1, 5]
]

for nums in test_cases:
    winner = solve(nums)
    print(f"Array: {nums} ? Winner: {winner}")

Array: [1, 2, 3, 4, 5] ? Winner: Amal
Array: [5, 4, 3, 2, 1] ? Winner: Bimal
Array: [3, 1, 4, 1, 5] ? Winner: Amal

Conclusion

The solution counts the number of left-to-right maximums in the array. If this count is even, Amal wins; otherwise, Bimal wins. This approach has O(n) time complexity and O(1) space complexity.