Amazon Interview Question for Software Developers


Country: India




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A greedy approach can work here. The idea is that we keep track of how many incomplete stacks there are to the left of position i (shortage array). Note that shortage[-1] - shortage[i] also gives us the information on the number of incomplete stacks to the right of position i. Then for every stack where we have some extra coins (>1) we calculate how many needs to be moved left / right based on our knowledge of incomplete stacks:

def findMinMovesToRedistributeCoins(coins):
    shortage = [0 for _ in coins]
    missing = 0
    for i,x in enumerate(coins):
        if not x:
            missing += 1
        shortage[i] = missing
    ans = 0
    while shortage[-1]:
        for i,x in enumerate(coins):
            extra = x - 1
            if extra > 0 and i > 0 and shortage[i-1] > 0:
                # move stack left
                d = min(extra, shortage[i-1])
                coins[i] -= d
                if not coins[i-1]:
                    for j in range(i-1, len(coins)):
                        shortage[j] -= 1
                coins[i-1] += d
                # print coins
                ans += 1
                extra -= d
            if extra > 0 and i+1<len(coins) and shortage[-1] != shortage[i]:
                # move stack right
                d = min(extra, shortage[-1] - shortage[i])
                coins[i] -= d
                if not coins[i+1]:
                    for j in range(i+1, len(coins)):
                        shortage[j] -= 1
                coins[i+1] += d
                # print coins
                ans += 1
    return ans

Test driver:

print findMinMovesToRedistributeCoins([0, 2])
[1, 1]
1

print findMinMovesToRedistributeCoins([0, 3, 1, 0, 3, 0, 0])
[1, 2, 1, 0, 3, 0, 0]
[1, 1, 2, 0, 3, 0, 0]
[1, 1, 1, 1, 3, 0, 0]
[1, 1, 1, 1, 1, 2, 0]
[1, 1, 1, 1, 1, 1, 1]
5

- adr October 07, 2018 | Flag Reply


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