Description#
You are given an integer array arr of length n that represents a permutation of the integers in the range [0, n - 1].
We split arr into some number of chunks (i.e., partitions), and individually sort each chunk. After concatenating them, the result should equal the sorted array.
Return the largest number of chunks we can make to sort the array.
Example 1:
Input: arr = [4,3,2,1,0]
Output: 1
Explanation:
Splitting into two or more chunks will not return the required result.
For example, splitting into [4, 3], [2, 1, 0] will result in [3, 4, 0, 1, 2], which isn't sorted.
Example 2:
Input: arr = [1,0,2,3,4]
Output: 4
Explanation:
We can split into two chunks, such as [1, 0], [2, 3, 4].
However, splitting into [1, 0], [2], [3], [4] is the highest number of chunks possible.
Constraints:
n == arr.length1 <= n <= 100 <= arr[i] < n- All the elements of
arr are unique.
Solutions#
Solution 1#
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| class Solution:
def maxChunksToSorted(self, arr: List[int]) -> int:
mx = ans = 0
for i, v in enumerate(arr):
mx = max(mx, v)
if i == mx:
ans += 1
return ans
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| class Solution {
public int maxChunksToSorted(int[] arr) {
int ans = 0, mx = 0;
for (int i = 0; i < arr.length; ++i) {
mx = Math.max(mx, arr[i]);
if (i == mx) {
++ans;
}
}
return ans;
}
}
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| class Solution {
public:
int maxChunksToSorted(vector<int>& arr) {
int ans = 0, mx = 0;
for (int i = 0; i < arr.size(); ++i) {
mx = max(mx, arr[i]);
ans += i == mx;
}
return ans;
}
};
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| func maxChunksToSorted(arr []int) int {
ans, mx := 0, 0
for i, v := range arr {
mx = max(mx, v)
if i == mx {
ans++
}
}
return ans
}
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| function maxChunksToSorted(arr: number[]): number {
const n = arr.length;
let ans = 0;
let max = 0;
for (let i = 0; i < n; i++) {
max = Math.max(arr[i], max);
if (max == i) {
ans++;
}
}
return ans;
}
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| impl Solution {
pub fn max_chunks_to_sorted(arr: Vec<i32>) -> i32 {
let mut res = 0;
let mut max = 0;
for i in 0..arr.len() {
max = max.max(arr[i]);
if max == (i as i32) {
res += 1;
}
}
res
}
}
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| #define max(a, b) (((a) > (b)) ? (a) : (b))
int maxChunksToSorted(int* arr, int arrSize) {
int res = 0;
int mx = -1;
for (int i = 0; i < arrSize; i++) {
mx = max(mx, arr[i]);
if (mx == i) {
res++;
}
}
return res;
}
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Solution 2#
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| class Solution:
def maxChunksToSorted(self, arr: List[int]) -> int:
stk = []
for v in arr:
if not stk or v >= stk[-1]:
stk.append(v)
else:
mx = stk.pop()
while stk and stk[-1] > v:
stk.pop()
stk.append(mx)
return len(stk)
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| class Solution {
public int maxChunksToSorted(int[] arr) {
Deque<Integer> stk = new ArrayDeque<>();
for (int v : arr) {
if (stk.isEmpty() || v >= stk.peek()) {
stk.push(v);
} else {
int mx = stk.pop();
while (!stk.isEmpty() && stk.peek() > v) {
stk.pop();
}
stk.push(mx);
}
}
return stk.size();
}
}
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| class Solution {
public:
int maxChunksToSorted(vector<int>& arr) {
stack<int> stk;
for (int v : arr) {
if (stk.empty() || v >= stk.top()) {
stk.push(v);
} else {
int mx = stk.top();
stk.pop();
while (!stk.empty() && stk.top() > v) {
stk.pop();
}
stk.push(mx);
}
}
return stk.size();
}
};
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| func maxChunksToSorted(arr []int) int {
stk := []int{}
for _, v := range arr {
if len(stk) == 0 || v >= stk[len(stk)-1] {
stk = append(stk, v)
} else {
mx := stk[len(stk)-1]
stk = stk[:len(stk)-1]
for len(stk) > 0 && stk[len(stk)-1] > v {
stk = stk[:len(stk)-1]
}
stk = append(stk, mx)
}
}
return len(stk)
}
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