# Solution For Max Chunks To Make Sorted

The Max Chunks To Make Sorted problem on LeetCode asks us to find the maximum number of chunks that can be formed to sort an array of distinct integers.

To solve this problem, we can use a simple approach where we iterate over the array and keep track of the maximum value of the elements seen so far. We can then compare this maximum value with the current index and increment the count of chunks if the maximum value is equal to the current index.

Here is the step-by-step approach to solve this problem:

1. Initialize a variable called count to 0 which will store the number of chunks.
2. Initialize a variable called max_val to 0 which will store the maximum value seen so far.
3. Loop through the array and do the following:
a. Update the max_val variable by taking the maximum of the current element and the max_val.
b. If the max_val is equal to the current index i, increment the count variable and reset the max_val to 0.
4. Return the count variable, which will be the maximum number of chunks that can be formed.

Here is the Python code for the above approach:

`def maxChunksToSorted(arr): count = 0 max_val = 0 for i in range(len(arr)): max_val = max(max_val, arr[i]) if max_val == i: count += 1 max_val = 0 return count`

Let’s consider an example to understand how the above code works:

Example:
`arr = [4, 3, 2, 1, 0]`
Initially, count = 0 and max_val = 0.

When i = 0, arr[i] = 4, max_val = 4 which is not equal to i, so we don’t increment the count.

When i = 1, arr[i] = 3, max_val = 4 which is not equal to i, so we don’t increment the count.

When i = 2, arr[i] = 2, max_val = 4 which is not equal to i, so we don’t increment the count.

When i = 3, arr[i] = 1, max_val = 4 which is not equal to i, so we don’t increment the count.

When i = 4, arr[i] = 0, max_val = 4 which is equal to i, so we increment the count to 1 and reset the max_val to 0.

Therefore, the maximum number of chunks that can be formed is 1.

The time complexity of this approach is O(n) where n is the length of the array.

## Step by Step Implementation For Max Chunks To Make Sorted

```class Solution {
public int maxChunksToSorted(int[] arr) {
//This problem can be solved using the concept of maximum subarray sum
//We can keep track of the maximum value seen so far and once the maximum value seen so far becomes equal
//to the current index value, we can increment the count of chunks
int maxSeen = Integer.MIN_VALUE, count = 0;
for(int i=0; idef maxChunksToSorted(arr):

# initialize the number of chunks
# with the length of the array
num_chunks = len(arr)

# initialize the maximum value seen
# so far with the first element of the array
max_val_seen_so_far = arr[0]

# start iterating from the second element
# of the array till the end
for i in range(1, len(arr)):

# update the maximum value seen so far
# if the current element is greater
max_val_seen_so_far = max(max_val_seen_so_far, arr[i])

# if the maximum value seen so far is equal
# to the current index, it means that
# till this index, all the elements are
# sorted. So, increase the number of chunks
if max_val_seen_so_far == i:
num_chunks += 1

# return the number of chunks
return num_chunksYou can use the following approach to solve this problem:

1. Iterate through the array and keep track of the maximum value seen so far.
2. Whenever you encounter a value that is equal to the maximum value seen so far, you know that all values up to that point can be placed in a separate chunk.
3. Repeat until you reach the end of the array.

For example, given the array [5, 4, 3, 2, 1], you would iterate through it as follows:

1. The maximum value seen so far is 5.
2. You encounter a 4, which is less than the maximum value seen so far, so you continue.
3. You encounter a 3, which is less than the maximum value seen so far, so you continue.
4. You encounter a 2, which is less than the maximum value seen so far, so you continue.
5. You encounter a 1, which is less than the maximum value seen so far, so you continue.
6. You reach the end of the array, so you know that all values can be placed in separate chunks.

Thus, the answer is 6.int maxChunksToSorted(vector& arr) {

// maxChunksToSorted will take in an array of integers and return
// the maximum number of chunks that can be made such that each
// chunk is sorted in ascending order.

// We will use a greedy approach to solve this problem.
// We will iterate through the array and keep track of the maximum
// element seen so far. Whenever we see an element that is greater
// than or equal to the maximum element seen so far, we know that
// all of the elements up to that point can be placed in a single
// chunk. We can then update the maximum element seen so far and
// continue iterating.

int max_element = arr[0]; // Maximum element seen so far
int num_chunks = 0; // Number of chunks that can be made

// Iterate through the array
for (int i = 0; i < arr.size(); i++) {
// Update the maximum element seen so far, if necessary
if (arr[i] > max_element) {
max_element = arr[i];
}
// If the current element is equal to the maximum element seen so far,
// we can place it in a chunk.
if (arr[i] == max_element) {
num_chunks++;
}
}

return num_chunks;
}public int MaxChunksToSorted(int[] arr) { // keep track of the max value we've seen so far int max = 0; // keep track of the number of chunks int chunks = 0; // iterate through the array for (int i = 0; i < arr.Length; i++) { // update the max value max = Math.Max(max, arr[i]); // if the current index is equal to the max value, // then we know we can make a chunk if (i == max) { chunks++; } } return chunks; }```
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