Problem Description

You are given a 0-indexed integer array nums.

You can perform any number of operations on this array. In each operation, you:

1. Select a subarray (a contiguous non-empty sequence of elements)
2. Replace that subarray with a single element equal to the sum of all elements in that subarray

For example, if nums = [1,3,5,6] and you select the subarray [3,5], the array becomes [1,8,6] (the subarray [3,5] is replaced by their sum 8).

Your goal is to find the maximum length of a non-decreasing array that can be obtained after performing any number of these operations.

A non-decreasing array means each element is greater than or equal to the previous element (i.e., nums[i] <= nums[i+1] for all valid indices).

Examples:

• nums = [5,2,2]: The original array is not non-decreasing. You can:

  • Replace [2,2] with [4] to get [5,4] - still not non-decreasing
  • Replace [5,2] with [7] to get [7,2] - still not non-decreasing
  • Replace [5,2,2] with [9] to get [9] - this is non-decreasing with length 1
  • Maximum achievable length is 1

• nums = [1,2,3,4]: Already non-decreasing, so maximum length is 4

• nums = [4,3,2,6]: Replace [3,2] with [5] to get [4,5,6] which is non-decreasing with length 3

The challenge is to strategically choose which subarrays to merge to create the longest possible non-decreasing sequence.

Intuition

The key insight is that when we merge subarrays into their sum, we're essentially creating segments where each segment has a value (the sum of its elements). To maximize the length of the final non-decreasing array, we want to create as many segments as possible while maintaining the non-decreasing property.

Think about it this way: if we have a segment ending at position i with sum S, the next segment starting from position i+1 must have a sum at least S to maintain the non-decreasing property. This means we want to find the shortest possible next segment whose sum is at least S.

This leads us to a dynamic programming approach where:

• f[i] represents the maximum number of segments we can create using the first i elements
• For each position i, we need to track what was the last position where we ended a segment that allows us to create valid segments up to position i

The crucial observation is that if we ended a segment at position pre[i], and the sum from pre[i] to i is sum1, then the next segment must have sum at least sum1. To find where the next segment can end, we need to find the smallest index j where the sum from i to j is at least sum1.

Using prefix sums s[i] to quickly calculate range sums:

• Sum from position a to b is s[b] - s[a]
• If the last segment ended at pre[i] with sum s[i] - s[pre[i]], we need the next segment to have sum at least this value
• We're looking for the smallest j where s[j] - s[i] >= s[i] - s[pre[i]]
• Rearranging: s[j] >= 2*s[i] - s[pre[i]]

This is why the solution uses bisect_left to find the earliest position where we can end the next segment, and maintains pre[j] to track the best previous segment ending position for each index.

Learn more about Stack, Queue, Binary Search, Dynamic Programming, Monotonic Queue and Monotonic Stack patterns.

Solution Approach

The solution uses dynamic programming with prefix sums and binary search to efficiently find the maximum number of segments (which equals the maximum length of the final non-decreasing array).

Data Structures Used:

• s: Prefix sum array where s[i] = sum of nums[0] to nums[i-1]
• f[i]: Maximum number of segments achievable using the first i elements
• pre[i]: The best ending position of the last segment before position i

Algorithm Steps:

1. Initialize prefix sums: Create s using accumulate(nums, initial=0) so that s[i] stores the sum of elements from index 0 to i-1.

2. Dynamic Programming Loop: For each position i from 1 to n:

   a. Inherit the best previous position: pre[i] = max(pre[i], pre[i-1])

   • This ensures we're using the optimal ending position from previous computations

   b. Calculate maximum segments: f[i] = f[pre[i]] + 1

   • We can form one more segment than what we had at position pre[i]
   • The new segment spans from pre[i] to i

   c. Find next valid position: Use binary search to find the smallest index j where we can end the next segment

   • The sum of the current segment is s[i] - s[pre[i]]
   • The next segment starting at i must have sum ≥ s[i] - s[pre[i]]
   • We need s[j] - s[i] >= s[i] - s[pre[i]], which gives us s[j] >= 2*s[i] - s[pre[i]]
   • j = bisect_left(s, 2*s[i] - s[pre[i]])

   d. Update future position: pre[j] = i

   • Record that position i is a good ending position for creating segments up to position j

3. Return Result: f[n] contains the maximum number of segments using all n elements.

Why this works:

• By maintaining pre[i], we always know the optimal way to partition elements up to position i
• The binary search ensures we find the earliest possible position to end the next segment while maintaining the non-decreasing property
• The DP recurrence f[i] = f[pre[i]] + 1 builds up the solution optimally from smaller subproblems

Time Complexity: O(n log n) due to the binary search operation for each of the n positions. Space Complexity: O(n) for the auxiliary arrays.

Example Walkthrough

Let's walk through the solution with nums = [4,3,2,6].

Step 1: Initialize prefix sums

• nums = [4,3,2,6]
• s = [0,4,7,9,15] (cumulative sums)
• n = 4
• f = [0,0,0,0,0] (max segments for each position)
• pre = [0,0,0,0,0] (best previous segment ending)

Step 2: Process each position

i = 1:

• pre[1] = max(0, 0) = 0
• f[1] = f[0] + 1 = 0 + 1 = 1 (one segment from 0 to 1)
• Current segment sum: s[1] - s[0] = 4 - 0 = 4
• Find next position j where sum ≥ 4:
  • Need s[j] >= 2*4 - 0 = 8
  • bisect_left(s, 8) returns 3 (since s[3] = 9 ≥ 8)
• pre[3] = 1 (position 1 is good for segments up to position 3)

i = 2:

• pre[2] = max(0, 0) = 0
• f[2] = f[0] + 1 = 1 (one segment from 0 to 2)
• Current segment sum: s[2] - s[0] = 7 - 0 = 7
• Find next position j where sum ≥ 7:
  • Need s[j] >= 2*7 - 0 = 14
  • bisect_left(s, 14) returns 4 (since s[4] = 15 ≥ 14)
• pre[4] = 2 (position 2 is good for segments up to position 4)

i = 3:

• pre[3] = max(1, 0) = 1 (from earlier update)
• f[3] = f[1] + 1 = 1 + 1 = 2 (two segments: [4] and [3,2])
• Current segment sum: s[3] - s[1] = 9 - 4 = 5
• Find next position j where sum ≥ 5:
  • Need s[j] >= 2*9 - 4 = 14
  • bisect_left(s, 14) returns 4
• pre[4] = max(2, 3) = 3 (update to better position)

i = 4:

• pre[4] = max(3, 0) = 3 (from earlier updates)
• f[4] = f[3] + 1 = 2 + 1 = 3 (three segments)
• Current segment sum: s[4] - s[3] = 15 - 9 = 6
• Find next position j where sum ≥ 6:
  • Need s[j] >= 2*15 - 9 = 21
  • bisect_left(s, 21) returns 5 (beyond array)
• No update needed (j > n)

Result: f[4] = 3

This means we can create 3 segments:

• Segment 1: [4] with sum 4
• Segment 2: [3,2] with sum 5
• Segment 3: [6] with sum 6

The final array [4,5,6] is non-decreasing with length 3.

Solution Implementation

Time and Space Complexity

Time Complexity: O(n log n)

The algorithm iterates through the array once with a loop from 1 to n, which takes O(n) time. Within each iteration:

• Accessing and updating arrays pre and f takes O(1) time
• The bisect_left operation performs binary search on the prefix sum array s, which takes O(log n) time

Since we perform the binary search operation for each of the n iterations, the overall time complexity is O(n) * O(log n) = O(n log n).

Space Complexity: O(n)

The algorithm uses several auxiliary arrays:

• s: prefix sum array of size n + 1
• f: DP array of size n + 1
• pre: array of size n + 2

Each array uses linear space proportional to the input size n. Therefore, the total space complexity is O(n) + O(n) + O(n) = O(n).

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Off-by-One Errors with Index Management

The most common pitfall in this solution is confusion around the indexing conventions, particularly with the last_segment_start array and how it relates to the prefix sums.

The Problem:

• The prefix sum array has length n+1 where prefix_sums[i] represents the sum of elements from index 0 to i-1
• The last_segment_start array needs to handle positions beyond n (hence size n+2)
• When using bisect_left, the returned index might be n+1, which would cause an index out of bounds error if not handled properly

Example of the pitfall:

# Incorrect: Array too small
last_segment_start = [0] * (n + 1)  # This will cause IndexError
# When bisect_left returns n+1, last_segment_start[n+1] will fail

Solution: Always ensure the last_segment_start array has size n+2 to accommodate all possible positions returned by bisect_left.

2. Incorrect Prefix Sum Calculation

The Problem: Using the wrong initial value or incorrect accumulation can lead to wrong segment sums.

Example of the pitfall:

# Incorrect: No initial value
prefix_sums = list(accumulate(nums))  # Length is n, not n+1
# This makes prefix_sums[0] = nums[0] instead of 0

Solution: Always use accumulate(nums, initial=0) to ensure prefix_sums[0] = 0 and the array has length n+1.

3. Misunderstanding the Binary Search Target

The Problem: The binary search looks for the minimum position where we can end the next segment while maintaining the non-decreasing property. The calculation 2 * prefix_sums[i] - prefix_sums[last_segment_start[i]] might seem arbitrary.

Example of the pitfall:

# Incorrect understanding might lead to:
next_position = bisect_left(prefix_sums, prefix_sums[i] + segment_sum, i+1)
# Adding a start parameter to bisect_left, which changes the behavior

Solution: Remember that we're searching for the first position j where:

• The sum from position i to j-1 is at least segment_sum
• This translates to: prefix_sums[j] >= prefix_sums[i] + segment_sum
• Which simplifies to: prefix_sums[j] >= 2 * prefix_sums[i] - prefix_sums[last_segment_start[i]]

4. Not Propagating the Best Previous Position

The Problem: Forgetting to update last_segment_start[i] with the maximum of itself and the previous position.

Example of the pitfall:

# Incorrect: Missing the max operation
for i in range(1, n + 1):
    # last_segment_start[i] = max(last_segment_start[i], last_segment_start[i - 1])  # Missing!
    dp[i] = dp[last_segment_start[i]] + 1  # Will use 0 instead of optimal position

Solution: Always include the line last_segment_start[i] = max(last_segment_start[i], last_segment_start[i - 1]) to ensure we're using the best ending position discovered so far.