1526. Minimum Number of Increments on Subarrays to Form a Target Array LeetCode Solution

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Table of Contents

1. Problem Statement
2. Complexity Analysis
3. Minimum Number of Increments on Subarrays to Form a Target Array solution in C++
4. Minimum Number of Increments on Subarrays to Form a Target Array solution in Java
5. Minimum Number of Increments on Subarrays to Form a Target Array solution in Python
6. Additional Resources

Problem Statement of Minimum Number of Increments on Subarrays to Form a Target Array

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

Complexity Analysis

• Time Complexity: O(n)
• Space Complexity: O(1)

1526. Minimum Number of Increments on Subarrays to Form a Target Array LeetCode Solution in C++

class Solution {
 public:
  int minNumberOperations(vector<int>& target) {
    int ans = target.front();

    for (int i = 1; i < target.size(); ++i)
      if (target[i] > target[i - 1])
        ans += target[i] - target[i - 1];

    return ans;
  }
};
/* code provided by PROGIEZ */

1526. Minimum Number of Increments on Subarrays to Form a Target Array LeetCode Solution in Java

class Solution {
  public int minNumberOperations(int[] target) {
    int ans = target[0];

    for (int i = 1; i < target.length; ++i)
      if (target[i] > target[i - 1])
        ans += target[i] - target[i - 1];

    return ans;
  }
}
// code provided by PROGIEZ

1526. Minimum Number of Increments on Subarrays to Form a Target Array LeetCode Solution in Python

class Solution:
  def minNumberOperations(self, target: list[int]) -> int:
    ans = target[0]

    for a, b in zip(target, target[1:]):
      if a < b:
        ans += b - a

    return ans
 # code by PROGIEZ

Additional Resources

• Explore all LeetCode problem solutions at Progiez here
• Explore all problems on LeetCode website here

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