LeetCode 1658. Minimum operand to reduce x to 0

1658. take x Reduced to 0 Minimum operands of

Give you an array of integers nums And an integer x . At each operation , You should remove the array nums Leftmost or rightmost element , And then from x Subtract the value of the element from the . Please note that , need modify Array for subsequent operations .

If you can x just Reduced to 0 , return Minimum operands ; otherwise , return -1 .

Example 1：

 Input ：nums = [1,1,4,2,3], x = 5
 Output ：2
 explain ： The best solution is to remove the last two elements , take  x  Reduced to  0 .

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
 explain ： The best solution is to remove the last three elements and the first two elements （ in total  5  operations ）, take  x  Reduced to  0 .

Tips ：

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

Two 、 Method 1

Hash + The prefix and

class Solution {
    
    public int minOperations(int[] nums, int x) {
    
        int len = nums.length;
        Map<Integer, Integer> map = new HashMap<>();
        int[] sum = new int[len + 1];
        map.put(0, 0);
        int temp = 0;
        for (int i = 1; i <= len; i++) {
    
            map.put(temp = sum[i - 1] + nums[i - 1], i);
            sum[i] = temp;
        }
        int target = sum[len] - x;
        if (target < 0) {
    
            return -1;
        }
        int res = Integer.MAX_VALUE;
        for (int i = len; i > 0 && sum[i] >= target; i--) {
    
            if (map.containsKey(temp = sum[i] - target)) {
    
                res = Math.min(res, len - i + map.get(temp));
            }
        }
        return res == Integer.MAX_VALUE ? -1 : res;
    }
}

Complexity analysis

• Time complexity ：O(n).

• Spatial complexity ：O(n).