feat: add solutions to lc problems: No.1658,2401 (doocs#2578)

* No.1658.Minimum Operations to Reduce X to Zero
* No.2401.Longest Nice Subarray

Author: yanglbme

solution/1600-1699/1658.Minimum Operations to Reduce X to Zero/README.md

@@ -1,20 +1,19 @@
-class Solution {
-public:
-    int minOperations(vector<int>& nums, int x) {
-        x = accumulate(nums.begin(), nums.end(), 0) - x;
-        unordered_map<int, int> vis{{0, -1}};
-        int n = nums.size();
-        int ans = 1 << 30;
-        for (int i = 0, s = 0; i < n; ++i) {
-            s += nums[i];
-            if (!vis.count(s)) {
-                vis[s] = i;
-            }
-            if (vis.count(s - x)) {
-                int j = vis[s - x];
-                ans = min(ans, n - (i - j));
-            }
-        }
-        return ans == 1 << 30 ? -1 : ans;
-    }
+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;
+    }
 };

solution/1600-1699/1658.Minimum Operations to Reduce X to Zero/README_EN.md

@@ -1,22 +1,21 @@
-func minOperations(nums []int, x int) int {
-	x = -x
-	for _, v := range nums {
-		x += v
-	}
-	vis := map[int]int{0: -1}
-	ans := 1 << 30
-	s, n := 0, len(nums)
-	for i, v := range nums {
-		s += v
-		if _, ok := vis[s]; !ok {
-			vis[s] = i
-		}
-		if j, ok := vis[s-x]; ok {
-			ans = min(ans, n-(i-j))
-		}
-	}
-	if ans == 1<<30 {
-		return -1
-	}
-	return ans
+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
 }

solution/1600-1699/1658.Minimum Operations to Reduce X to Zero/Solution.c

@@ -1,21 +1,20 @@
-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;
-    }
+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;
+    }
 }

solution/1600-1699/1658.Minimum Operations to Reduce X to Zero/Solution.cpp

@@ -1,14 +1,12 @@
-class Solution:
-    def minOperations(self, nums: List[int], x: int) -> int:
-        x = sum(nums) - x
-        vis = {0: -1}
-        ans = inf
-        s, n = 0, len(nums)
-        for i, v in enumerate(nums):
-            s += v
-            if s not in vis:
-                vis[s] = i
-            if s - x in vis:
-                j = vis[s - x]
-                ans = min(ans, n - (i - j))
-        return -1 if ans == inf else ans
+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

solution/1600-1699/1658.Minimum Operations to Reduce X to Zero/Solution.go

@@ -1,26 +1,25 @@
+use std::collections::HashMap;
+
 impl Solution {
     pub fn min_operations(nums: Vec<i32>, x: i32) -> i32 {
-        let n = nums.len();
-        let target = nums.iter().sum::<i32>() - x;
-        if target < 0 {
-            return -1;
-        }
-        let mut ans = i32::MAX;
-        let mut sum = 0;
-        let mut i = 0;
-        for j in 0..n {
-            sum += nums[j];
-            while sum > target {
-                sum -= nums[i];
-                i += 1;
+        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 sum == target {
-                ans = ans.min((n - 1 - (j - i)) as i32);
+            if let Some(&j) = vis.get(&(t - s)) {
+                mx = mx.max((i as i32) - j);
             }
         }
-        if ans == i32::MAX {
-            return -1;
+        if mx == -1 {
+            -1
+        } else {
+            (nums.len() as i32) - mx
         }
-        ans
     }
 }

solution/1600-1699/1658.Minimum Operations to Reduce X to Zero/Solution.java

@@ -1,18 +1,16 @@
 function minOperations(nums: number[], x: number): number {
-    x = nums.reduce((a, b) => a + b, 0) - x;
-    const vis = new Map();
-    vis.set(0, -1);
+    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;
-    let ans = 1 << 30;
-    for (let i = 0, s = 0; i < n; ++i) {
-        s += nums[i];
-        if (!vis.has(s)) {
-            vis.set(s, i);
+    for (let i = 0; i < n; ++i) {
+        t += nums[i];
+        if (!vis.has(t)) {
+            vis.set(t, i);
         }
-        if (vis.has(s - x)) {
-            const j = vis.get(s - x);
-            ans = Math.min(ans, n - (i - j));
+        if (vis.has(t - s)) {
+            mx = Math.max(mx, i - vis.get(t - s)!);
         }
     }
-    return ans == 1 << 30 ? -1 : ans;
+    return ~mx ? n - mx : -1;
 }

solution/1600-1699/1658.Minimum Operations to Reduce X to Zero/Solution.py

@@ -1,18 +1,18 @@
-class Solution {
-public:
-    int minOperations(vector<int>& nums, int x) {
-        x = accumulate(nums.begin(), nums.end(), 0) - x;
-        int n = nums.size();
-        int ans = 1 << 30;
-        for (int i = 0, j = 0, s = 0; i < n; ++i) {
-            s += nums[i];
-            while (j <= i && s > x) {
-                s -= nums[j++];
-            }
-            if (s == x) {
-                ans = min(ans, n - (i - j + 1));
-            }
-        }
-        return ans == 1 << 30 ? -1 : ans;
-    }
+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;
+    }
 };