2 k-length arrays DP solution

• Inspired by a solution in Best Time to Buy and Sell Stock III. The solution of this problem can only be got from sell k times, buy k times, sell k-1 times, buy k-1 times, ..., sell 1st time, buy 1st time. So we just need to iterate those possibilities.

``````public int maxProfit(int k, int[] prices) {
if (prices == null || k < 1) {
return 0;
}

if (prices.length <= k) { // just do greedy
return helper(prices);
}

final int[] sells = new int[k];
final int[] holds = new int[k];

for (int i = 0; i < holds.length; i++) {
holds[i] = Integer.MIN_VALUE;
}

for (int i = 0; i < prices.length; i++) {
for (int j = k - 1; j >= 0; j--) {
sells[j] = Math.max(sells[j], holds[j] + prices[i]);
int sell = j == 0 ? 0 : sells[j - 1];
holds[j] = Math.max(holds[j], sell - prices[i]);
}
}
return sells[k - 1];
}

private int helper(int[] prices) {
int max = 0;
for (int i = 0; i < prices.length - 1; i++) {
if (prices[i] < prices[i + 1]) {
max += prices[i + 1] - prices[i];
}
}
return max;
}
``````

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