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Minimum Operations to Reduce X to Zero

A curated StudyFlow practice problem with comprehensive editorial and multi-language solutions.

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Problem Statement

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 [2, 3] with sum 5 in 2 operations.

Example 2:

Input: nums = [5,6,7,8,9], x = 4
Output: -1
Explanation: There is no way to reduce x to 0 because all elements are greater than 4.

Example 3:

Input: nums = [3,2,20,1,1,3], x = 10
Output: 5
Explanation: Remove the 3 elements from the left [3, 2] and 2 elements from the right [1, 1, 3] (total sum 10, total 5 operations).

Constraints:

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

Prerequisites

Before attempting this problem, you should be comfortable with:

  • Sliding Window on positive integer arrays
  • Prefix sums and algebraic reduction
  • Two-pointer boundary convergence

What You'll Learn

Key principles and algorithmic techniques you will master from this problem:

  • Master the technique of inverting boundary prefix/suffix problems into middle subarray sliding window problems.
  • Understand why strictly positive array elements guarantee sliding window monotonicity.
  • Handle edge cases like target < 0, target == 0, and x > totalSum.
  • Calculate minimum operations as n - maxSubarrayLength.

Pattern Recognition Trigger

How to instantly recognize this algorithmic pattern when encountering unseen interview questions:

Signals to Look For:

  • Picking elements from only the two ends (left and right) to reach a sum target.
  • All elements are strictly positive (nums[i] >= 1).
  • Inverting the boundary problem to a middle contiguous subarray problem.

Algorithmic Decision Tree:

  • If elements could be negative -> Hash Map Prefix Sum in O(N) time and O(N) space.
  • If elements are strictly positive -> Two-Pointer Sliding Window in O(N) time and O(1) space.
  • If target = totalSum - x is negative -> Return -1 immediately.

Connected Patterns:

Maximum Points You Can Obtain from CardsSubarray Sum Equals KMinimum Size Subarray Sum

1. Brute Force (Recursive Branching)

Intuition

Recursively explore picking the left element nums[left] or right element nums[right]. Return the minimum steps among valid branches.

Algorithm

  1. Step 1: In helper dfs(left, right, remX, steps):
  2. - If remX == 0, return steps.
  3. - If remX < 0 or left > right, return INF.
  4. - Recurse left: dfs(left + 1, right, remX - nums[left], steps + 1).
  5. - Recurse right: dfs(left, right - 1, remX - nums[right], steps + 1).
  6. - Return min(pickLeft, pickRight).
  7. Step 2: Return result or -1 if INF.

Prefix/Suffix Combinations (O(N^2) Time)

2 of 2 questions left today
Step 1 / 6
Trace timeline17% explored
Current operationSTEP 1 OF 6
Prefix [1, 1, 4] alone has sum 6, which overshoots x
Visual state
Values and pointers at this exact step
pointers
S
1
1
4
2
3
0
1
2
↑ prefix
3
4
Variables at this step
prefixLen
3
suffixLen
0
currentSum
6
x
5
minOps
null
Check / comparison
Prefix of length 3: 1 + 1 + 4 = 6 > 5
Decision / next transition
Sum exceeds x = 5. Invalid combination.
Why this step matters

Prefix [1, 1, 4] alone has sum 6, which overshoots x.

Step 1 / 6
C++ SOLUTIONCLEAN & OPTIMIZED
1#include <iostream>
2#include <vector>
3#include <algorithm>
4
5class Solution {
6private:
7 const int INF = 1e9;
8 int dfs(const std::vector<int>& nums, int left, int right, int x, int steps) {
9 if (x == 0) return steps;
10 if (x < 0 || left > right) return INF;
11
12 int pickLeft = dfs(nums, left + 1, right, x - nums[left], steps + 1);
13 int pickRight = dfs(nums, left, right - 1, x - nums[right], steps + 1);
14
15 return std::min(pickLeft, pickRight);
16 }
17
18public:
19 int minOperations(std::vector<int>& nums, int x) {
20 int res = dfs(nums, 0, (int)nums.size() - 1, x, 0);
21 return res >= INF ? -1 : res;
22 }
23};
24
25int main() {
26 Solution sol;
27 std::vector<int> nums = {1, 1, 4, 2, 3};
28 std::cout << "Min Operations: " << sol.minOperations(nums, 5) << std::endl;
29 return 0;
30}

Time & Space Complexity

  • Time complexity: O(N^2) — Time complexity for Brute Force (Recursive Branching).
  • Space complexity: O(1) — Space complexity for Brute Force (Recursive Branching).

Why Brute Force Doesn't Scale:

DFS takes O(2N) steps, which for N = 100,000 exceeds 2100000. Inverting to middle subarray sliding window reduces the entire problem to a single linear pass of 200,000 pointer operations.

For N = 100,000: DFS hangs indefinitely; Two-pointer sliding window completes in ~3 milliseconds.

2. Better (Hash Map Prefix Sums on Target Sum)

Intuition

Calculate target = totalSum - x. If target < 0, return -1. Use a hash map storing prefixSum -> index. If prefixSum - target exists in the map at index prev, subarray from prev + 1 to i has sum target. We maximize this subarray length.

Algorithm

  1. Step 1: Compute totalSum = sum(nums) and target = totalSum - x.
  2. Step 2: If target < 0, return -1. If target == 0, return n.
  3. Step 3: Initialize hash map map with map[0] = -1, currSum = 0, and maxLen = -1.
  4. Step 4: For each i from 0 to n - 1:
  5. - currSum += nums[i].
  6. - If currSum - target exists in map, maxLen = max(maxLen, i - map[currSum - target]).
  7. - If currSum not in map, map[currSum] = i.
  8. Step 5: Return maxLen == -1 ? -1 : n - maxLen.

Prefix Sums + Hash Map of Suffix Sums (O(N) Space)

2 of 2 questions left today
Step 1 / 6
Trace timeline17% explored
Current operationSTEP 1 OF 6
Suffix map allows instant complement lookups
Visual state
Values and pointers at this exact step
pointers
S
1
1
4
2
3
0
↑ init
1
2
3
4
Variables at this step
suffixMap
{"0":0,"3":1,"5":2}
x
5
minOps
null
Check / comparison
Build suffix sum map from right to left: { 0:0, 3:1, 5:2, 9:3, 10:4, 11:5 }
Decision / next transition
Map maps required suffix sum to number of operations.
Why this step matters

Suffix map allows instant complement lookups.

Step 1 / 6
C++ SOLUTIONCLEAN & OPTIMIZED
1#include <iostream>
2#include <vector>
3#include <numeric>
4#include <unordered_map>
5#include <algorithm>
6
7class Solution {
8public:
9 int minOperations(std::vector<int>& nums, int x) {
10 long long totalSum = std::accumulate(nums.begin(), nums.end(), 0LL);
11 long long target = totalSum - x;
12 int n = nums.size();
13
14 if (target < 0) return -1;
15 if (target == 0) return n;
16
17 std::unordered_map<long long, int> map;
18 map[0] = -1;
19 long long currSum = 0;
20 int maxLen = -1;
21
22 for (int i = 0; i < n; i++) {
23 currSum += nums[i];
24 if (map.count(currSum - target)) {
25 maxLen = std::max(maxLen, i - map[currSum - target]);
26 }
27 if (!map.count(currSum)) {
28 map[currSum] = i;
29 }
30 }
31 return maxLen == -1 ? -1 : n - maxLen;
32 }
33};
34
35int main() {
36 Solution sol;
37 std::vector<int> nums = {1, 1, 4, 2, 3};
38 std::cout << "Better Min Ops: " << sol.minOperations(nums, 5) << std::endl;
39 return 0;
40}

Time & Space Complexity

  • Time complexity: O(N log N) — Time complexity for Better (Hash Map Prefix Sums on Target Sum).
  • Space complexity: O(N) — Space complexity for Better (Hash Map Prefix Sums on Target Sum).

3. Optimal (Sliding Window on Target Sum)

Intuition

Since nums[i] >= 1, the sum of any window [left...right] increases monotonically when right advances and decreases monotonically when left advances. We expand right and shrink left while currentSum > target. When currentSum == target, we update maxLen. Finally, return n - maxLen.

Algorithm

  1. Step 1: Compute totalSum = sum(nums) and target = totalSum - x.
  2. Step 2: If target < 0, return -1. If target == 0, return n.
  3. Step 3: Initialize left = 0, currSum = 0, and maxLen = -1.
  4. Step 4: For right from 0 to nums.length - 1:
  5. - currSum += nums[right].
  6. - While currSum > target and left <= right:
  7. - currSum -= nums[left].
  8. - left++.
  9. - If currSum == target:
  10. - maxLen = max(maxLen, right - left + 1).
  11. Step 5: Return maxLen == -1 ? -1 : n - maxLen.

Inverted Sliding Window: Longest Subarray of Sum (Total - X)

2 of 2 questions left today
Step 1 / 6
Trace timeline17% explored
Current operationSTEP 1 OF 6
Retaining middle [1, 1, 4] leaves n - 3 = 5 - 3 = 2 operations on the ends
Visual state
Values and pointers at this exact step
pointers
S
1
1
4
2
3
0
↑ L
1
2
↑ R
3
4
Variables at this step
L
0
R
2
currentSum
6
targetSum
6
maxLen
3
minOps
2
Check / comparison
Total sum = 11. Target middle sum = 11 - 5 = 6. Expand window: 1 + 1 + 4 = 6
Decision / next transition
Window [0..2] has sum 6 == target! Length = 3. maxLen = 3.
Why this step matters

Retaining middle [1, 1, 4] leaves n - 3 = 5 - 3 = 2 operations on the ends.

Step 1 / 6
C++ SOLUTIONCLEAN & OPTIMIZED
1#include <iostream>
2#include <vector>
3#include <numeric>
4#include <algorithm>
5
6class Solution {
7public:
8 int minOperations(std::vector<int>& nums, int x) {
9 long long totalSum = std::accumulate(nums.begin(), nums.end(), 0LL);
10 long long target = totalSum - x;
11 int n = nums.size();
12
13 if (target < 0) return -1;
14 if (target == 0) return n;
15
16 int left = 0;
17 long long currSum = 0;
18 int maxLen = -1;
19
20 for (int right = 0; right < n; right++) {
21 currSum += nums[right];
22
23 while (currSum > target && left <= right) {
24 currSum -= nums[left];
25 left++;
26 }
27
28 if (currSum == target) {
29 maxLen = std::max(maxLen, right - left + 1);
30 }
31 }
32 return maxLen == -1 ? -1 : n - maxLen;
33 }
34};
35
36int main() {
37 Solution sol;
38 std::vector<int> nums = {1, 1, 4, 2, 3};
39 std::cout << "Optimal Min Ops: " << sol.minOperations(nums, 5) << std::endl;
40 return 0;
41}

Time & Space Complexity

  • Time complexity: O(N) — Time complexity for Optimal (Sliding Window on Target Sum).
  • Space complexity: O(1) — Space complexity for Optimal (Sliding Window on Target Sum).

Approach Comparison Matrix

Side-by-side comparison of time, space, and algorithmic trade-offs across each approach:

ApproachTimeSpaceCore IdeaTrade-off / Bottleneck
1.Brute Force (Recursive Branching / DFS)O(2^N)O(N) call stackAt each step, choose to take either the leftmost or rightmost element and recurse with remaining x.Exponential recursion tree branching.
2.Better (Hash Map Prefix Sums on Target Sum)O(N)O(N)Compute target = totalSum - x. Use a hash map storing prefix sum -> index to find max-length subarray with sum target.Hash map lookups and memory allocations.
3.Optimal (Sliding Window on Target Sum)OptimalO(N)O(1)Compute target = totalSum - x. Use two pointers left and right to find longest subarray with sum target.Single pass scan bound.

Common Pitfalls & Crucial Edge Cases

Trying to greedily pick larger element from left vs right

Greedy choice fails (e.g. taking 5 on left might block a 2+3 on right that perfectly equals x).

Forgetting `target < 0` check

If x > totalSum, target = totalSum - x is negative. A sliding window on positive integers will fail or loop infinitely if target < 0 is not caught early.

Returning `maxLen` instead of `n - maxLen`

The problem asks for the number of removed elements, not the length of the remaining subarray.

Crucial Edge Cases to Validate:

  • x equals totalSum (x = 11, totalSum = 11): target = 0. All elements must be removed; returns n.
  • x > totalSum (x = 50, totalSum = 11): target = -39 < 0. Returns -1 immediately.
  • No valid combination sums to x (nums = [5, 6, 7, 8, 9], x = 4): maxLen remains -1; returns -1.
  • Removing only from one end (nums = [5, 2, 3], x = 5): Subarray [2, 3] has sum 5, leaving 5 at index 0 (1 operation).

Interview Tips & Best Practices

How to communicate your thought process and address follow-ups in a live technical interview:

  1. Explain the problem inversion clearly: "Minimizing prefix + suffix elements summing to x is equivalent to maximizing middle subarray length summing to totalSum - x."
  2. State why the two-pointer approach works in O(1) space (all nums[i] are positive).
  3. Explicitly handle the early returns: if (target < 0) return -1; and if (target == 0) return n;.

Quick Comprehension Check

Test your intuitive understanding of the core concept in 30 seconds:

1.Why can we invert the problem of removing elements from both ends into a contiguous subarray problem?

2.Why does the Two-Pointer Sliding Window work in O(N) time without requiring a hash map?

3.What should the algorithm return if x > sum(nums)?

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