Simple dp solution with explanation
Intuition\nLet\'s start with a simple problem with one segment. This problem could be solved with a sliding window. To solve the problem with 2 segments we can 
# Intuition\nLet\'s start with a simple problem with one segment. This problem could be solved with a sliding window. To solve the problem with 2 segments we can use dynamic programming.\n\n# Approach\nLet\'s define a dp[k] as the max segment in the first k elements. We can iterate an array and for every, i build a max possible segment ending on the ith element. Imagine for particular i we have have segment starting on j. In this case, segment length would be `i - j + 1`. dp[i+1] - max possible segment in the array up to index i. We can calculate `dp[i+1] = max(dp[i], i + j - 1)`. If we have some valid segment from j to i, then a max of 2 segments including this one will be a max segment in the first j elements + length of current segment - `dp[j] + (i - j + 1)`. The final answer for this problem will be `max(dp[j] + (i - j + 1))` for every possible i.\n\n# Complexity\n- Time complexity: O(N)\n\n- Space complexity: O(N)\n\n# Code\n```\n/**\n * @param {number[]} prizePositions\n * @param {number} k\n * @return {number}\n */\nvar maximizeWin = function(prizePositions, k) {\n    const n = prizePositions.length;\n    let dp = new Array(n + 1).fill(0);\n    let j = 0;\n    let res = 0;\n    for(let i = 0; i < n; i++) {\n        while(prizePositions[i] > prizePositions[j] + k){\n            j++;\n        }\n        let len = i - j + 1;\n        dp[i + 1] = Math.max(dp[i], len);\n        res = Math.max(res, dp[j] + len);\n    }\n    \n    return res;\n};\n\n\n```
['Dynamic Programming', 'JavaScript']