Create 2025-maximum-number-of-ways-to-partition-an-array.js

Author: everthis

2025-maximum-number-of-ways-to-partition-an-array.js

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+/**
+ * @param {number[]} nums
+ * @param {number} k
+ * @return {number}
+ */
+const waysToPartition = function (nums, k) {
+  const n = nums.length
+  const pref = Array(n).fill(0),
+    suff = Array(n).fill(0)
+  pref[0] = nums[0]
+  suff[n - 1] = nums[n - 1]
+  for (let i = 1; i < n; ++i) {
+    pref[i] = pref[i - 1] + nums[i]
+    suff[n - 1 - i] = suff[n - i] + nums[n - 1 - i]
+  }
+  let ans = 0
+  const left = {},
+    right = {}
+
+  for (let i = 0; i < n - 1; ++i) {
+    const delta = pref[i] - suff[i + 1]
+    if (right[delta] == null) right[delta] = 0
+    right[delta]++
+  }
+
+  if (right[0]) ans = right[0]
+  for (let i = 0; i < n; ++i) {
+    //find the number of pivot indexes when nums[i] is changed to k
+    let curr = 0,
+      diff = k - nums[i]
+    if (left[diff]) curr += left[diff]
+    if (right[-diff]) curr += right[-diff]
+
+    //update answer
+    ans = Math.max(ans, curr)
+
+    //transfer the current element from right to left
+    if (i < n - 1) {
+      let dd = pref[i] - suff[i + 1]
+      if(left[dd] == null) left[dd] = 0
+      if(right[dd] == null) right[dd] = 0
+      left[dd]++
+      right[dd]--
+      if (right[dd] == 0) delete right[dd]
+    }
+  }
+  return ans
+}