You are given two arrays, nums1 and nums2, of the same length . For each index , you must choose exactly one number: either nums1[i] (which you treat as positive) or nums2[i] (which you treat as negative). You need to find the number of ways to choose one number from each index for a subarray nums[L...R] such that the sum of the chosen numbers is exactly 0.
Adobe uses this "Hard" problem to test your expertise in Dynamic Programming and the "Meet-in-the-middle" or "Target Sum" concepts. It evaluates your ability to handle large search spaces by identifying that the sum of a subarray can be broken down into transitions from previous indices. It’s a test of whether you can optimize or brute-force logic into a more efficient DP approach.
This is a Dynamic Programming problem on subarrays. Since each subarray can start at any index , the total number of subarrays is . For each possible start, you can use DP to track the number of ways to reach a specific sum at each subsequent index. The state dp[i][current_sum] would store the number of ways to reach current_sum using elements from L to i. To avoid start points, you can use a single DP that effectively "starts" a new potential sum at every index.
nums1 = [1, 2], nums2 = [1, 2]
Possible choices for each index: {+1, -1}, {+2, -2}.
Subarrays:
nums1[i] or -nums2[i]).One major mistake is misunderstanding the choice rule—you must pick one number for every index in the chosen range . Another is failing to use a Map for the DP state, which is necessary because the possible sums can be large and sparse. Many candidates also forget to apply a modulo if the number of ways is large.
This is a variation of the "Subset Sum" or "Target Sum" problem. Practice these classic DP problems to become comfortable with "ways to reach sum " logic. For subarray problems, remember that you can "reset" the DP or start a new sum at every index.
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