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Minimum Operations to Reduce X to Zero

Med
#1522Time: O(2^n), where n is the length of `nums`. For each element in the current subarray, we branch into two possibilities, leading to an exponential number of function calls.Space: O(n), where n is the length of `nums`. This is due to the maximum depth of the recursion call stack.1 company
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Prompt

You are given an integer array nums and an integer x. In one operation, you can either remove the leftmost or the rightmost element from the array nums and subtract its value from x. Note that this modifies the array for future operations.

Return the minimum number of operations to reduce x to exactly 0 if it is possible, otherwise, return -1.

 

Example 1:

Input: nums = [1,1,4,2,3], x = 5
Output: 2
Explanation: The optimal solution is to remove the last two elements to reduce x to zero.

Example 2:

Input: nums = [5,6,7,8,9], x = 4
Output: -1

Example 3:

Input: nums = [3,2,20,1,1,3], x = 10
Output: 5
Explanation: The optimal solution is to remove the last three elements and the first two elements (5 operations in total) to reduce x to zero.

 

Constraints:

  • 1 <= nums.length <= 105
  • 1 <= nums[i] <= 104
  • 1 <= x <= 109

Approaches

3 approaches with complexity analysis and trade-offs.

This approach directly simulates the problem statement. We can define a recursive function that explores every possible sequence of removing elements from either the left or the right end of the array. At each step, we have two choices: remove the leftmost element or the rightmost element. We explore both paths and continue until x becomes zero or it's impossible to reach zero.

Algorithm

  • Define a recursive function solve(nums, x, left, right).
  • Base Case 1: If x is 0, a solution is found for the subproblem. Return 0 operations.
  • Base Case 2: If x becomes negative or the pointers cross (left > right), this path is invalid. Return a very large number (infinity) to signify failure.
  • Recursive Step: Explore two choices:
    1. Remove the leftmost element: res1 = 1 + solve(nums, x - nums[left], left + 1, right).
    2. Remove the rightmost element: res2 = 1 + solve(nums, x - nums[right], left, right - 1).
  • Return the minimum of res1 and res2.
  • The initial call is solve(nums, x, 0, nums.length - 1). If the result is infinity, it means no solution was found, so return -1.

Walkthrough

The core idea is to use a recursive function, say solve(left, right, x), which returns the minimum operations to make x zero using the subarray nums[left...right]. This method explores the entire decision tree of removals.

  • Base Cases:

    • If x == 0, we've successfully reduced it to zero. No more operations are needed for this subproblem, so we return 0.
    • If x < 0 or left > right, it's impossible to reach the target from this state. We return a value indicating impossibility, like Integer.MAX_VALUE.

  • Recursive Step:

    • We make two recursive calls representing the two possible moves:
      1. Remove nums[left]: 1 + solve(left + 1, right, x - nums[left])
      2. Remove nums[right]: 1 + solve(left, right - 1, x - nums[right])
    • The function returns the minimum of these two results.

This approach is very slow because it recomputes results for the same subproblems multiple times and has an exponential number of paths to explore. Memoization is difficult because the state depends on (left, right, x), and x can be large.

// Note: This solution is for demonstration and will time out on larger test cases.class Solution {    public int minOperations(int[] nums, int x) {        int result = solve(nums, x, 0, nums.length - 1);        return result >= 1_000_000_000 ? -1 : result;    }     private int solve(int[] nums, int x, int left, int right) {        if (x == 0) {            return 0;        }        if (x < 0 || left > right) {            return 1_000_000_000; // Using a large number for infinity        }         int takeLeft = 1 + solve(nums, x - nums[left], left + 1, right);        int takeRight = 1 + solve(nums, x - nums[right], left, right - 1);         return Math.min(takeLeft, takeRight);    }}

Complexity

Time

O(2^n), where n is the length of `nums`. For each element in the current subarray, we branch into two possibilities, leading to an exponential number of function calls.

Space

O(n), where n is the length of `nums`. This is due to the maximum depth of the recursion call stack.

Trade-offs

Pros

  • Simple to understand as it directly models the process described in the problem.

Cons

  • Extremely inefficient and will result in a 'Time Limit Exceeded' error for all but the smallest inputs.

  • Performs a lot of redundant computations for the same subproblems.

Solutions

class Solution {public  int minOperations(int[] nums, int x) {    x = -x;    for (int v : nums) {      x += v;    }    Map<Integer, Integer> vis = new HashMap<>();    vis.put(0, -1);    int n = nums.length;    int ans = 1 << 30;    for (int i = 0, s = 0; i < n; ++i) {      s += nums[i];      vis.putIfAbsent(s, i);      if (vis.containsKey(s - x)) {        int j = vis.get(s - x);        ans = Math.min(ans, n - (i - j));      }    }    return ans == 1 << 30 ? -1 : ans;  }}

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