Asked in: AMAZON
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All Testcases Passed ✔
Solution
def findGrowthCoefficient(cityStart, cityEnd, m):
# Write your code here
h = {}
for i,j in zip(cityStart, cityEnd):
// ... rest of solution available after purchaseExplanation
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To solve this problem, you need to compute a special metric called the "growth coefficient" for a 1-dimensional country where cities are numbered from 1 to m. The country has n warehouses, and each warehouse has a specific delivery range from cityStart[i] to cityEnd[i], inclusive. A key function, F(x), represents how many warehouses can deliver to the x-th city. The final result — the growth coefficient — is calculated by applying the XOR operation across all F(x) values for cities 1 through m.
To begin solving this, think about how to efficiently compute F(x) for all cities from 1 to m. A brute-force approach would be to iterate through each city and for each city, loop through all warehouses to check if it is within their delivery range. This would result in O(m * n) time complexity, which is acceptable only for small values of m and n. However, if m and n can be large (e.g., up to 10^5), then this approach becomes inefficient.
So, the first step in designing an efficient solution is to understand how to determine how many ranges (warehouses) cover each city, in an optimized way. A good way to approach this is by using a "difference array" technique, commonly used in range update problems.
The idea is to treat the delivery coverage as a prefix sum problem. You initialize an auxiliary array of size m+2 (to handle boundaries), and for each warehouse delivery range from cityStart[i] to cityEnd[i], you increment the start index by 1 and decrement the index after the end by 1. This way, you mark the beginning and the end of a range, but you delay the actual counting. After processing all delivery ranges in this way, a simple prefix sum pass over the array gives you the exact number of warehouses covering each city.
Once you have computed the array F where F[x] gives the number of warehouses delivering to city x, you then compute the growth coefficient by applying the XOR operation across all values from F[1] to F[m].
The XOR operation itself is straightforward — it’s associative and commutative, so you can process it in a single linear scan. Start with an initial result of zero and XOR it with F[1], then F[2], and so on up to F[m].
Let’s revisit the example provided to validate the approach. Suppose:
- cityStart = [1, 1, 2]
- cityEnd = [1, 3, 4]
- m = 4
Using the difference array idea:
- For warehouse 1: update position 1 (+1) and position 2 (-1)
- For warehouse 2: update position 1 (+1) and position 4 (-1)
- For warehouse 3: update position 2 (+1) and position 5 (-1)
After processing these ranges, do a prefix sum:
- At city 1: sum = 2 (warehouse 1 and 2)
- At city 2: sum = 2 (warehouse 2 and 3)
- At city 3: sum = 1 (only warehouse 2)
- At city 4: sum = 1 (only warehouse 3)
So F = [2, 2, 1, 1]
Now compute XOR:
2 ^ 2 = 0
0 ^ 1 = 1
1 ^ 1 = 0
The final growth coefficient is 0.
This example shows that you can handle overlapping delivery ranges efficiently and determine the exact number of warehouses covering each city using an auxiliary array and prefix sum.
To summarize your approach:
1. Initialize an auxiliary array of size m+2 with zeros.
2. For each warehouse range, increment the start position and decrement the position after the end.
3. Compute the prefix sum over the auxiliary array to get F(x) for all cities.
4. Iterate through F[1] to F[m] and compute the cumulative XOR.
5. Return this XOR value as the final growth coefficient.
This method ensures an efficient and scalable solution to the problem, avoiding unnecessary nested loops and utilizing prefix sums to handle range coverage efficiently.
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