1526. Minimum Number of Increments on Subarrays to Form a Target Array
Description
You are given an integer array target
. You have an integer array initial
of the same size as target
with all elements initially zeros.
In one operation you can choose any subarray from initial
and increment each value by one.
Return the minimum number of operations to form a target
array from initial
.
The test cases are generated so that the answer fits in a 32-bit integer.
Example 1:
Input: target = [1,2,3,2,1] Output: 3 Explanation: We need at least 3 operations to form the target array from the initial array. [0,0,0,0,0] increment 1 from index 0 to 4 (inclusive). [1,1,1,1,1] increment 1 from index 1 to 3 (inclusive). [1,2,2,2,1] increment 1 at index 2. [1,2,3,2,1] target array is formed.
Example 2:
Input: target = [3,1,1,2] Output: 4 Explanation: [0,0,0,0] -> [1,1,1,1] -> [1,1,1,2] -> [2,1,1,2] -> [3,1,1,2]
Example 3:
Input: target = [3,1,5,4,2] Output: 7 Explanation: [0,0,0,0,0] -> [1,1,1,1,1] -> [2,1,1,1,1] -> [3,1,1,1,1] -> [3,1,2,2,2] -> [3,1,3,3,2] -> [3,1,4,4,2] -> [3,1,5,4,2].
Constraints:
1 <= target.length <= 105
1 <= target[i] <= 105
Solutions
Solution 1: Dynamic Programming
We define $f[i]$ as the minimum number of operations required to obtain $target[0,..i]$, initially setting $f[0] = target[0]$.
For $target[i]$, if $target[i] \leq target[i-1]$, then $f[i] = f[i-1]$; otherwise, $f[i] = f[i-1] + target[i] - target[i-1]$.
The final answer is $f[n-1]$.
We notice that $f[i]$ only depends on $f[i-1]$, so we can maintain the operation count using just one variable.
The time complexity is $O(n)$, where $n$ is the length of the array $target$. The space complexity is $O(1)$.
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