Partition an array such into maximum increasing segments

• Difficulty Level : Hard
• Last Updated : 10 May, 2021

We are given an array of N integers, we need to partition the array into segments such that every element of a segment is greater than every element of previous segment. In other words, if we sort these individual segments, the whole array becomes sorted. We need to find a valid partition with maximum number of subarrays.
Examples:

arr[] = {3 1 2 4 100 7 9}
Output : 3
You should partition the array into the following subarrays: (3, 1, 2), (4) and (100, 7, 9).
Input : arr[] = {2 1 2 3 3 4 3}
Output : 5

Approach A greedy approach works really well as it leads to the following algorithm. Find the shortest prefix such that all the elements in the prefix are less than or equal to the elements in the rest of the array.
Consider this prefix as the first subarray of the partition. Recursively call the same algorithm on the rest of the array. Implementation wise, we can preprocess an array of max values for each prefix and another array of min values for each suffix. This way we can easily perform a check to see if a given prefix is a viable candidate for a partition subarray.
Below is the implementation of above approach:

C++

 // C++ program to divide into maximum number of segments #include using namespace std;     // Returns the maximum number of sorted subarrays    // in a valid partition    int sorted_partitions(int arr[],int n)    {                 int right_min[n + 1];        right_min[n] = INT8_MAX;        for (int i = n - 1; i >= 0; i--) {            right_min[i] = min(right_min[i + 1], arr[i]);        }         // Finding the shortest prefix such that all the elements        // in the prefix are less than or equal to the elements        // in the rest of the array.        int partitions = 0;        for (int current_max = arr, i = 0; i < n; i++) {            current_max = max(current_max, arr[i]);                       // if current max is less than the right prefix min,            // we increase number of partitions.            if (current_max <= right_min[i + 1])                partitions++;        }         return partitions;    }     // Driver code    int main()    {        int arr[] = { 3, 1, 2, 4, 100, 7, 9 };        // Find minimum value from right for every index        int n = sizeof(arr)/sizeof(arr);                 int ans = sorted_partitions(arr,n);        cout << ans << endl;        return 0;             // This code is contributed by ANKITRAI1    }

Java

 // Java program to divide into maximum number of segmentsimport java.util.Arrays; class GFG {     // Returns the maximum number of sorted subarrays    // in a valid partition    static int sorted_partitions(int arr[])    {        // Find minimum value from right for every index        int n = arr.length;        int[] right_min = new int[n + 1];        right_min[n] = Integer.MAX_VALUE;        for (int i = n - 1; i >= 0; i--) {            right_min[i] = Math.min(right_min[i + 1], arr[i]);        }         // Finding the shortest prefix such that all the elements        // in the prefix are less than or equal to the elements        // in the rest of the array.        int partitions = 0;        for (int current_max = arr, i = 0; i < n; i++) {            current_max = Math.max(current_max, arr[i]);                       // if current max is less than the right prefix min,            // we increase number of partitions.            if (current_max <= right_min[i + 1])                partitions++;        }         return partitions;    }     // Driver code    public static void main(String[] args)    {        int[] arr = new int[] { 3, 1, 2, 4, 100, 7, 9 };        int ans = sorted_partitions(arr);        System.out.println(ans);    }}

Python 3

 # Python 3 program to divide into# maximum number of segmentsimport sys # Returns the maximum number of sorted# subarrays in a valid partitiondef sorted_partitions( arr, n):         right_min =  * (n + 1)    right_min[n] = sys.maxsize    for i in range(n - 1, -1, -1):        right_min[i] = min(right_min[i + 1], arr[i])         # Finding the shortest prefix such that    # all the elements in the prefix are less    # than or equal to the elements in the    # rest of the array.    partitions = 0    current_max = arr    for i in range(n):        current_max = max(current_max, arr[i])                 # if current max is less than the right        # prefix min, we increase number of partitions.        if (current_max <= right_min[i + 1]):            partitions += 1     return partitions # Driver codearr = [ 3, 1, 2, 4, 100, 7, 9 ] # Find minimum value from right# for every indexn = len(arr)ans = sorted_partitions(arr, n)print(ans)     # This code is contributed by ita_c

C#

 // C# program to divide into maximum number of segmentsusing System; class GFG {     // Returns the maximum number of sorted subarrays    // in a valid partition    static int sorted_partitions(int []arr)    {        // Find minimum value from right for every index        int n = arr.Length;        int[] right_min = new int[n + 1];        right_min[n] =  int.MaxValue;        for (int i = n - 1; i >= 0; i--) {            right_min[i] = Math.Min(right_min[i + 1], arr[i]);        }         // Finding the shortest prefix such that all the elements        // in the prefix are less than or equal to the elements        // in the rest of the array.        int partitions = 0;        for (int current_max = arr, i = 0; i < n; i++) {            current_max = Math.Max(current_max, arr[i]);                     // if current max is less than the right prefix min,            // we increase number of partitions.            if (current_max <= right_min[i + 1])                partitions++;        }         return partitions;    }     // Driver code    public static void Main()    {        int[] arr = { 3, 1, 2, 4, 100, 7, 9 };        int ans = sorted_partitions(arr);        Console.WriteLine(ans);    }}// This code is contributed by anuj_67..

PHP

 = 0; \$i--)    {        \$right_min[\$i] = min(\$right_min[\$i + 1],                                        \$arr[\$i]);    }     // Finding the shortest prefix such    // that all the elements in the prefix    // are less than or equal to the elements    // in the rest of the array.    \$partitions = 0;    for (\$current_max = \$arr,                        \$i = 0; \$i < \$n; \$i++)    {        \$current_max = max(\$current_max, \$arr[\$i]);                 // if current max is less than the        // right prefix min, we increase        // number of partitions.        if (\$current_max <= \$right_min[\$i + 1])            \$partitions++;    }     return \$partitions;} // Driver code\$arr = array( 3, 1, 2, 4, 100, 7, 9 ); // Find minimum value from// right for every index\$n = sizeof(\$arr); \$ans = sorted_partitions(\$arr, \$n);echo \$ans, "\n"; // This code is contributed by ajit?>

Javascript


Output:
3

My Personal Notes arrow_drop_up