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authorjack <[email protected]>2024-06-14 13:13:15 -0600
committerjack <[email protected]>2024-06-14 13:13:15 -0600
commitd343a3b18dbaca9297d4f7b2280c4a16f0a5c8aa (patch)
tree96df922dae2a5073d9a50ea70c7aae6aa37e3ebe /partition_array_max_sum.cpp
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+// 1043. Partition Array for Maximum Sum
+#include "leetcode.h"
+
+/*
+ * given an integeer array 'arr', partition the array into contiguous subarrays
+ * of length at most 'k'. after partitioning, each subarray has their values
+ * changed to become the maximum value of that subarray. return the largest sum
+ * of the given array after partitioning. test cases are generated so that
+ * answer fits in a 32-bit integer
+ */
+
+class Solution {
+public:
+ int maxSumAfterPartitioning(vector<int> &arr, int k) {
+ int n = arr.size();
+ vector<int> dp(n + 1);
+ for (int i = 1; i <= n; ++i) {
+ int curr = 0, best = 0;
+ for (int j = 1; j <= k && i - j >= 0; ++j) {
+ curr = max(curr, arr[i - j]);
+ best = max(best, dp[i - j] + curr * j);
+ }
+ dp[i] = best;
+ }
+ return dp[n];
+ }
+};
+
+int main() {
+ Solution obj;
+ vector<int> a1 = {1, 15, 7, 9, 2, 5, 10},
+ a2 = {1, 4, 1, 5, 7, 3, 6, 1, 9, 9, 3}, a3 = {1};
+ printf("%d\n", obj.maxSumAfterPartitioning(a1, 3)); // expect: 84
+ printf("%d\n", obj.maxSumAfterPartitioning(a2, 4)); // expect: 83
+ printf("%d\n", obj.maxSumAfterPartitioning(a3, 1)); // expect: 1
+}