Description
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 <= 1051 <= nums[i] <= 1041 <= x <= 109
Solutions
Solution 1: Hash Table + Prefix Sum
According to the problem description, we need to remove elements from both ends of the array nums so that the sum of the removed elements equals x, and the number of removed elements is minimized. We can transform the problem into: find the longest consecutive subarray in the array nums such that the sum of the subarray s = ∑_{i=0}n nums[i] - x. In this way, we can transform the problem into finding the length mx of the longest consecutive subarray in the array nums with a sum of s, and the answer is n - mx.
We initialize mx = -1, and then use a hash table vis to store the prefix sum, where the key is the prefix sum and the value is the index corresponding to the prefix sum.
Traverse the array nums, for the current element nums[i], calculate the prefix sum t, if t is not in the hash table, add t to the hash table; if t - s is in the hash table, update mx = max(mx, i - vis[t - s]).
Finally, if mx = -1, return -1, otherwise return n - mx.
The time complexity is O(n), and the space complexity is O(n). Where n is the length of the array nums.
PythonJavaC++GoTypeScriptRust
class Solution:
def minOperations(self, nums: List[int], x: int) -> int:
s = sum(nums) - x
vis = {0: -1}
mx, t = -1, 0
for i, v in enumerate(nums):
t += v
if t not in vis:
vis[t] = i
if t - s in vis:
mx = max(mx, i - vis[t - s])
return -1 if mx == -1 else len(nums) - mx(code-box)
class Solution {
public int minOperations(int[] nums, int x) {
int s = -x;
for (int v : nums) {
s += v;
}
Map<Integer, Integer> vis = new HashMap<>();
vis.put(0, -1);
int mx = -1, t = 0;
int n = nums.length;
for (int i = 0; i < n; ++i) {
t += nums[i];
vis.putIfAbsent(t, i);
if (vis.containsKey(t - s)) {
mx = Math.max(mx, i - vis.get(t - s));
}
}
return mx == -1 ? -1 : n - mx;
}
}(code-box)
class Solution {
public:
int minOperations(vector<int>& nums, int x) {
int s = accumulate(nums.begin(), nums.end(), 0) - x;
unordered_map<int, int> vis = {{0, -1}};
int mx = -1, t = 0;
int n = nums.size();
for (int i = 0; i < n; ++i) {
t += nums[i];
if (!vis.contains(t)) {
vis[t] = i;
}
if (vis.contains(t - s)) {
mx = max(mx, i - vis[t - s]);
}
}
return mx == -1 ? -1 : n - mx;
}
};(code-box)
func minOperations(nums []int, x int) int {
s := -x
for _, v := range nums {
s += v
}
vis := map[int]int{0: -1}
mx, t := -1, 0
for i, v := range nums {
t += v
if _, ok := vis[t]; !ok {
vis[t] = i
}
if j, ok := vis[t-s]; ok {
mx = max(mx, i-j)
}
}
if mx == -1 {
return -1
}
return len(nums) - mx
}(code-box)
function minOperations(nums: number[], x: number): number {
const s = nums.reduce((acc, cur) => acc + cur, -x);
const vis: Map<number, number> = new Map([[0, -1]]);
let [mx, t] = [-1, 0];
const n = nums.length;
for (let i = 0; i < n; ++i) {
t += nums[i];
if (!vis.has(t)) {
vis.set(t, i);
}
if (vis.has(t - s)) {
mx = Math.max(mx, i - vis.get(t - s)!);
}
}
return ~mx ? n - mx : -1;
}(code-box)
use std::collections::HashMap;
impl Solution {
pub fn min_operations(nums: Vec<i32>, x: i32) -> i32 {
let s = nums.iter().sum::<i32>() - x;
let mut vis: HashMap<i32, i32> = HashMap::new();
vis.insert(0, -1);
let mut mx = -1;
let mut t = 0;
for (i, v) in nums.iter().enumerate() {
t += v;
if !vis.contains_key(&t) {
vis.insert(t, i as i32);
}
if let Some(&j) = vis.get(&(t - s)) {
mx = mx.max((i as i32) - j);
}
}
if mx == -1 {
-1
} else {
(nums.len() as i32) - mx
}
}
}(code-box)
Solution 2: Two Pointers
Based on the analysis of Solution 1, we need to find the length mx of the longest consecutive subarray in the array nums with a sum of s. Since all elements in the array nums are positive integers, the prefix sum of the array will only increase monotonically, so we can use two pointers to solve this problem.
We initialize pointer j = 0, prefix sum t = 0, and the length of the longest consecutive subarray mx = -1.
Traverse the array nums, for the current element nums[i], calculate the prefix sum t += nums[i]. If t > s, then move the pointer j until t ≤ s. If t = s, then update mx = max(mx, i - j + 1).
Finally, if mx = -1, return -1, otherwise return n - mx.
The time complexity is O(n), where n is the length of the array nums. The space complexity is O(1).
PythonJavaC++GoTypeScriptRust
class Solution:
def minOperations(self, nums: List[int], x: int) -> int:
s = sum(nums) - x
j = t = 0
mx = -1
for i, x in enumerate(nums):
t += x
while j <= i and t > s:
t -= nums[j]
j += 1
if t == s:
mx = max(mx, i - j + 1)
return -1 if mx == -1 else len(nums) - mx(code-box)
class Solution {
public int minOperations(int[] nums, int x) {
int s = -x;
for (int v : nums) {
s += v;
}
int mx = -1, t = 0;
int n = nums.length;
for (int i = 0, j = 0; i < n; ++i) {
t += nums[i];
while (j <= i && t > s) {
t -= nums[j++];
}
if (t == s) {
mx = Math.max(mx, i - j + 1);
}
}
return mx == -1 ? -1 : n - mx;
}
}(code-box)
class Solution {
public:
int minOperations(vector<int>& nums, int x) {
int s = accumulate(nums.begin(), nums.end(), 0) - x;
int mx = -1, t = 0;
int n = nums.size();
for (int i = 0, j = 0; i < n; ++i) {
t += nums[i];
while (j <= i && t > s) {
t -= nums[j++];
}
if (t == s) {
mx = max(mx, i - j + 1);
}
}
return mx == -1 ? -1 : n - mx;
}
};(code-box)
func minOperations(nums []int, x int) int {
s := -x
for _, v := range nums {
s += v
}
mx, t, j := -1, 0, 0
for i, v := range nums {
t += v
for ; j <= i && t > s; j++ {
t -= nums[j]
}
if t == s {
mx = max(mx, i-j+1)
}
}
if mx == -1 {
return -1
}
return len(nums) - mx
}(code-box)
function minOperations(nums: number[], x: number): number {
const s = nums.reduce((acc, cur) => acc + cur, -x);
let [mx, t] = [-1, 0];
const n = nums.length;
for (let i = 0, j = 0; i < n; ++i) {
t += nums[i];
while (t > s) {
t -= nums[j++];
}
if (t === s) {
mx = Math.max(mx, i - j + 1);
}
}
return ~mx ? n - mx : -1;
}(code-box)
impl Solution {
pub fn min_operations(nums: Vec<i32>, x: i32) -> i32 {
let s: i32 = nums.iter().sum::<i32>() - x;
let mut j: usize = 0;
let mut t: i32 = 0;
let mut mx: i32 = -1;
for (i, &v) in nums.iter().enumerate() {
t += v;
while j <= i && t > s {
t -= nums[j];
j += 1;
}
if t == s {
mx = mx.max((i - j + 1) as i32);
}
}
if mx == -1 {
-1
} else {
(nums.len() as i32) - mx
}
}
}(code-box)
