3229. Minimum Operations to Make Array Equal to Target LeetCode Solution
In this guide, you will get 3229. Minimum Operations to Make Array Equal to Target LeetCode Solution with the best time and space complexity. The solution to Minimum Operations to Make Array Equal to Target problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.
Problem Statement of Minimum Operations to Make Array Equal to Target
You are given two positive integer arrays nums and target, of the same length.
In a single operation, you can select any subarray of nums and increment each element within that subarray by 1 or decrement each element within that subarray by 1.
Return the minimum number of operations required to make nums equal to the array target.
Example 1:
Input: nums = [3,5,1,2], target = [4,6,2,4]
Output: 2
Explanation:
We will perform the following operations to make nums equal to target:
– Increment nums[0..3] by 1, nums = [4,6,2,3].
– Increment nums[3..3] by 1, nums = [4,6,2,4].
Example 2:
Input: nums = [1,3,2], target = [2,1,4]
Output: 5
Explanation:
We will perform the following operations to make nums equal to target:
– Increment nums[0..0] by 1, nums = [2,3,2].
– Decrement nums[1..1] by 1, nums = [2,2,2].
– Decrement nums[1..1] by 1, nums = [2,1,2].
– Increment nums[2..2] by 1, nums = [2,1,3].
– Increment nums[2..2] by 1, nums = [2,1,4].
3229. Minimum Operations to Make Array Equal to Target LeetCode Solution in C++
class Solution {
public:
// Similar to 1526. Minimum Number of Increments on Subarrays to Form a Target
// Array
long long minimumOperations(vector<int>& nums, vector<int>& target) {
long ans = abs(nums[0] - target[0]);
for (int i = 1; i < nums.size(); ++i) {
const int currDiff = target[i] - nums[i];
const int prevDiff = target[i - 1] - nums[i - 1];
if (currDiff >= 0 && prevDiff >= 0)
ans += max(0, currDiff - prevDiff);
else if (currDiff <= 0 && prevDiff <= 0)
ans += max(0, abs(currDiff) - abs(prevDiff));
else
ans += abs(currDiff);
}
return ans;
}
}; /* code provided by PROGIEZ */
3229. Minimum Operations to Make Array Equal to Target LeetCode Solution in Java
class Solution {
// Similar to 1526. Minimum Number of Increments on Subarrays to Form a Target Array
public long minimumOperations(int[] nums, int[] target) {
long ans = Math.abs((long) nums[0] - target[0]);
for (int i = 1; i < nums.length; ++i) {
final int currDiff = target[i] - nums[i];
final int prevDiff = target[i - 1] - nums[i - 1];
if (currDiff >= 0 && prevDiff >= 0)
ans += Math.max(0, currDiff - prevDiff);
else if (currDiff <= 0 && prevDiff <= 0)
ans += Math.max(0, Math.abs((long) currDiff) - Math.abs((long) prevDiff));
else
ans += Math.abs((long) currDiff);
}
return ans;
}
} // code provided by PROGIEZ
3229. Minimum Operations to Make Array Equal to Target LeetCode Solution in Python
class Solution:
# Similar to 1526. Minimum Number of Increments on Subarrays to Form a Target Array
def minimumOperations(self, nums: list[int], target: list[int]) -> int:
ans = abs(nums[0] - target[0])
for (prevNum, prevTarget), (currNum, currTarget) in (
itertools.pairwise(zip(nums, target))
):
currDiff = currTarget - currNum
prevDiff = prevTarget - prevNum
if currDiff >= 0 and prevDiff >= 0:
ans += max(0, currDiff - prevDiff)
elif currDiff <= 0 and prevDiff <= 0:
ans += max(0, abs(currDiff) - abs(prevDiff))
else:
ans += abs(currDiff)
return ans # code by PROGIEZ
