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dp with a size equal to the length of nums and fill it with -1, representing uncomputed states.solve that takes the current index i and the array nums.i is greater than or equal to the length of nums minus 2, return 0, as there are no more subarrays of size 3 to consider.dp[i] is not -1, return dp[i] as the already computed result.i), the next element (i+1), or the element after that (i+2).dp[i] and return it.nums and replace each element with the maximum of k - nums[i] and 0. This represents the minimum increments needed to make the single element beautiful.solve function starting from index 0, 1, and 2, and return the minimum of these three calls.