i (0-based) holds exactly i + 1 numbers, and pyramid[i][j] is the fee for stepping on cell j of that row.
You start at the apex pyramid[0][0] and go down one row per step until you reach the bottom. From cell j you may step to cell j or j + 1 of the row below. The cost of the descent is the sum of every number on the way, the apex and the bottom cell included.
Return the minimum cost of a descent.
Try to use only O(n) extra memory, where n is the number of rows.pyramid=[[3], [7, 1], [2, 9, 4], [8, 5, 6, 1]] → 9
The cheapest is 3 → 1 → 4 → 1, totalling 9. Greed doesn't always pay: going through 7 and 2 already costs 3 + 7 + 2 + 5 = 17.
pyramid=[[-2], [5, -6]] → -8
From the apex step down to −6: −2 + (−6) = −8.
pyramid=[[4]] → 4
pyramid=[[3], [7, 1], [2, 9, 4], [8, 5, 6, 1]]
9
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