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Given an array nums, start at index 0 and jump only forward to reach n-1, where a jump i→j yields score (j−i)*nums[i]; maximize the total score. This reduces to a DP dp[j] = max_i(dp[i] - i*nums[i] + j*nums[i]) and can be optimized from O(n^2) to O(n log n) (or O(n)) using a max line container / convex-hull-trick or Li Chao tree for n up to 1e5.
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Leetcode 3282. Reach End of Array With Max Score
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Source snapshot: Nov 12, 2026
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