Maximum sum of lengths of non-overlapping subarrays with k as the max element. - GeeksforGeeks

19 Apr, 2022
Find the maximum sum of lengths of non-overlapping subarrays (contiguous elements) with k as the maximum element. Examples: 
Input : arr[] = {2, 1, 4, 9, 2, 3, 8, 3, 4} 
        k = 4
Output : 5
{2, 1, 4} => Length = 3
{3, 4} => Length = 2
So, 3 + 2 = 5 is the answer

Input : arr[] = {1, 2, 3, 2, 3, 4, 1} 
        k = 4
Output : 7
{1, 2, 3, 2, 3, 4, 1} => Length = 7

Input : arr = {4, 5, 7, 1, 2, 9, 8, 4, 3, 1}
        k = 4
Ans = 4
{4} => Length = 1
{4, 3, 1} => Length = 3
So, 1 + 3 = 4 is the answer
question source : https://www.geeksforgeeks.org/amazon-interview-experience-set-376-campus-internship/
Algorithm : 
Traverse the array starting from first element
   Take a loop and keep on incrementing count 
   If element is less than equal to k
       if array element is equal to k, then mark
       a flag
   
   If flag is marked, add this count to answer
   
   Take another loop and traverse the array 
   till element is greater than k
return ans
C++
Java
Python3
C#
PHP
Javascript
// CPP program to calculate max sum lengths of// non overlapping contiguous subarrays with k as// max element#include <bits/stdc++.h>using namespace std;  // Returns max sum of lengths with maximum element// as kint calculateMaxSumLength(int arr[], int n, int k){    int ans = 0; // final sum of lengths      // number of elements in current subarray    int count = 0;      // variable for checking if k appeared in subarray    int flag = 0;      for (int i = 0; i < n;) {        count = 0;        flag = 0;          // count the number of elements which are        // less than equal to k        while (arr[i] <= k && i < n) {            count++;            if (arr[i] == k)                flag = 1;            i++;        }          // if current element appeared in current        // subarray add count to sumLength        if (flag == 1)            ans += count;             // skip the array elements which are        // greater than k        while (arr[i] > k && i < n)            i++;        }    return ans;}  // driver programint main(){    int arr[] = { 4, 5, 7, 1, 2, 9, 8, 4, 3, 1 };    int size = sizeof(arr) / sizeof(arr[0]);    int k = 4;    int ans = calculateMaxSumLength(arr, size, k);    cout << "Max Length :: " << ans << endl;    return 0;}
// A Java program to calculate max sum lengths of// non overlapping contiguous subarrays with k as// max elementpublic class GFG{    // Returns max sum of lengths with maximum element    // as k    static int calculateMaxSumLength(int arr[], int n, int k) {        int ans = 0; // final sum of lengths         // number of elements in current subarray        int count = 0;         // variable for checking if k appeared in subarray        int flag = 0;         for (int i = 0; i < n;) {            count = 0;            flag = 0;             // count the number of elements which are            // less than equal to k            while (i < n && arr[i] <= k) {                count++;                if (arr[i] == k)                    flag = 1;                i++;            }             // if current element appeared in current            // subarray add count to sumLength            if (flag == 1)                ans += count;             // skip the array elements which are            // greater than k            while (i < n && arr[i] > k)                i++;        }        return ans;    }     // driver program to test above method    public static void main(String[] args) {         int arr[] = { 4, 5, 7, 1, 2, 9, 8, 4, 3, 1 };        int size = arr.length;        int k = 4;        int ans = calculateMaxSumLength(arr, size, k);        System.out.println("Max Length :: " + ans);    }}// This code is contributed by Sumit Ghosh
# Python program to calculate max sum lengths of non# overlapping contiguous subarrays with k as max element # Returns max sum of lengths with max elements as kdef calculateMaxSumLength(arr, n, k):    ans = 0 # final sum of lengths    i=0    while i < n :                 # number of elements in current sub array        count = 0                 # Variable for checking if k appeared in the sub array        flag = 0                 # Count the number of elements which are        # less than or equal to k        while i < n and arr[i] <= k :            count = count + 1            if arr[i] == k:                flag = 1            i = i + 1                     # if current element appeared in current        # subarray and count to sumLength        if flag == 1:            ans = ans + count                     # skip the array elements which are greater than k        while i < n and arr[i] > k :            i = i + 1                  return ans     # Driver Programarr = [4, 5, 7, 1, 2, 9, 8, 4, 3, 1]size = len(arr)k = 4ans = calculateMaxSumLength(arr, size, k)print ("Max Length ::",ans) # Contributed by Rohit
// A C# program to calculate max// sum lengths of non overlapping// contiguous subarrays with k as// max elementusing System;class GFG {         // Returns max sum of lengths    // with maximum element as k    static int calculateMaxSumLength(int []arr,                                     int n,                                     int k)    {                 // final sum of lengths        int ans = 0;         // number of elements in        // current subarray        int count = 0;         // variable for checking if        // k appeared in subarray        int flag = 0;         for(int i = 0; i < n;)        {            count = 0;            flag = 0;             // count the number of            // elements which are            // less than equal to k            while (i < n && arr[i] <= k)            {                count++;                if (arr[i] == k)                    flag = 1;                i++;            }             // if current element            // appeared in current            // subarray add count            // to sumLength            if (flag == 1)                ans += count;             // skip the array            // elements which are            // greater than k            while (i < n && arr[i] > k)                i++;        }        return ans;    }     // Driver Code    public static void Main()    {        int []arr = {4, 5, 7, 1, 2, 9, 8, 4, 3, 1};        int size = arr.Length;        int k = 4;        int ans = calculateMaxSumLength(arr, size, k);        Console.WriteLine("Max Length :: " + ans);    }} // This code is contributed by anuj_67.
<?php// PHP program to calculate max sum lengths// of non overlapping contiguous subarrays// with k as max element // Returns max sum of lengths with maximum// element as kfunction calculateMaxSumLength(&$arr, $n, $k){    $ans = 0; // final sum of lengths     // number of elements in current subarray    $count = 0;     // variable for checking if k    // appeared in subarray    $flag = 0;     for ($i = 0; $i < $n    {        $count = 0;        $flag = 0;         // count the number of elements which        // are less than equal to k        while ($arr[$i] <= $k && $i < $n)        {            $count++;            if ($arr[$i] == $k)                $flag = 1;            $i++;        }         // if current element appeared in current        // subarray add count to sumLength        if ($flag == 1)            $ans += $count;         // skip the array elements which are        // greater than k        while ($arr[$i] > $k && $i < $n)            $i++;        }    return $ans;} // Driver Code$arr = array( 4, 5, 7, 1, 2,                 9, 8, 4, 3, 1 );$size = sizeof($arr);$k = 4;$ans = calculateMaxSumLength($arr, $size, $k);echo "Max Length :: " . $ans . "\n"; // This code is contributed by ita_c?>
<script>// A Javascript program to calculate max sum lengths of// non overlapping contiguous subarrays with k as// max element         // Returns max sum of lengths with maximum element    // as k    function calculateMaxSumLength(arr,n,k)    {        let ans = 0; // final sum of lengths          // number of elements in current subarray        let count = 0;          // variable for checking if k appeared in subarray        let flag = 0;          for (let i = 0; i < n;) {            count = 0;            flag = 0;              // count the number of elements which are            // less than equal to k            while (i < n && arr[i] <= k) {                count++;                if (arr[i] == k)                    flag = 1;                i++;            }              // if current element appeared in current            // subarray add count to sumLength            if (flag == 1)                ans += count;              // skip the array elements which are            // greater than k            while (i < n && arr[i] > k)                i++;        }        return ans;    }         // driver program to test above method    let arr=[4, 5, 7, 1, 2, 9, 8, 4, 3, 1];    let size = arr.length;    let k = 4;    let ans = calculateMaxSumLength(arr, size, k);    document.write("Max Length :: " + ans);         //This code is contributed by avanitrachhadiya2155     </script>
Max Length :: 4
Time Complexity : O(n) It may look like O(n2), but if you take a closer look, array is traversed only onceThis article is contributed by Mandeep Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
Algorithm:
Traverse the array from first element to last element
    if the element is less than k increment the count
    if the element is equals to k 
        if k is not found
            increment the count and mark flag as 1
        if k is found
            add the value of count to ans and mark count as 1
    if the element is greater than k
            if k is present in the subarray add the value of count to ans and
            assign value of count and flag variables as 0
finally check again if k value is found in subarray or not
    if k is found return sum of answer and count
    if not return ans
            
C++
Java
Python3
Javascript
// C++ program to find Maximum sum of lengths of// non-overlapping subarrays with k as the max element.#include <bits/stdc++.h>using namespace std;// Below function calculates the Maximum sum of lengths of// non-overlapping subarrays with k as the max element.int calculateMaxSumLength(int arr[], int n, int k){    // maximun sum of lengths    int ans = 0;    // number of elements in current subarray    int count = 0;    // falg variable for checking if k is present    // in current subarray or not    int flag = 0;    for (int i = 0; i < n; i++) {        // increment the count if element in arr is less        // than k        if (arr[i] < k) {            count++;        }        // if the element is equals to k        else if (arr[i] == k) {            if (flag == 0) {                count++;                flag = 1;            }            // if flag is 1, we can say k is already present            // in that subarray. So, add the value of count            // to ans variable and make the count value as 1            // because we found the k            else {                ans += count;                count = 1;            }        }        // if element in arr is grater than k        else {            // if k is present in the subarray            // add the value of count to ans variable            if (flag == 1) {                ans += count;            }            // assign value of count and flag variables as 0            count = 0;            flag = 0;        }    }    // Check again if k value is found in subarray    // if k is found, return sum of values of variables ans    // and count if k is not found, return value of variable    // ans.    if (flag == 1) {        return ans + count;    }    return ans;}// driver programint main(){    int arr[] = { 4, 5, 7, 1, 2, 9, 8, 4, 3, 1 };    int size = sizeof(arr) / sizeof(arr[0]);    int k = 4;    int ans = calculateMaxSumLength(arr, size, k);    cout << "Max Length :: " << ans << endl;    return 0;}// Contributed by Ravi Teja Kuchipudi
// JAVA program to find Maximum sum of lengths of// non-overlapping subarrays with k as the max element.class GFG {    // Below function calculates the Maximum sum of lengths    // of non-overlapping subarrays with k as the max    // element.    static int calculateMaxSumLength(int arr[], int n,                                     int k)    {        // maximun sum of lengths        int ans = 0;        // number of elements in current subarray        int count = 0;        // falg variable for checking if k is present        // in current subarray or not        int flag = 0;        for (int i = 0; i < n; i++) {            // increment the count if element in arr is less            // than k            if (arr[i] < k) {                count++;            }            // if the element is equals to k            else if (arr[i] == k) {                // if flag is equals to 0 then make flag                // variable value as 1 and increment the                // count.                if (flag == 0) {                    count++;                    flag = 1;                }                // if flag is 1, we can say k is already                // present in that subarray. So, add the                // value of count to ans variable and make                // the count value as 1 because we found the                // k                else {                    ans += count;                    count = 1;                }            }            // if element in arr is grater than k            else {                // if k is present in the subarray                // add the value of count to ans variable                if (flag == 1) {                    ans += count;                }                // assign value of count and flag variables                // as 0                count = 0;                flag = 0;            }        }        // Check again if k value is found in subarray        // if k is found, return sum of values of variables        // ans and count if k is not found, return value of        // variable ans.        if (flag == 1) {            return ans + count;        }        return ans;    }    // driver program to test above method    public static void main(String[] args)    {        int arr[] = { 4, 5, 7, 1, 2, 9, 8, 4, 3, 1 };        int size = arr.length;        int k = 4;        int ans = calculateMaxSumLength(arr, size, k);        System.out.println("Max Length :: " + ans);    }}// Contributed by Ravi Teja Kuchipudi
# program to find Maximum sum of lengths of# non-overlapping subarrays with k as the max element.  def calculateMaxSumLength(arr, n, k):    # maximun sum of lengths    ans = 0    # number of elements in current subarray    count = 0    # falg variable for checking if k is present in current subarray or not    flag = 0     for i in range(n):        # increment the count if element in arr is less than k        if arr[i] < k:            count = count+1        # if the element is equals to k        elif arr[i] == k:            # if flag is equals to 0 then make flag variable value as 1 and            # increment the count.            if flag == 0:                count = count + 1                flag = 1            # if flag is 1, we can say k is already present in that subarray.            # So, add the value of count to ans variable and            # make the count value as 1 because we found the k            else:                ans = ans + count                count = 1        # if element in arr is grater than k        else:            # if k is present in the subarray            # add the value of count to ans variable            if flag == 1:                ans = ans + count            # assign value of count and flag variables as 0            count = 0            flag = 0    # Check again if k value is found in subarray    # if k is found, return sum of values of variables ans and count    # if k is not found, return value of variable ans.    if flag == 1:        return ans + count    return ans  # Driver Programarr = [4, 5, 7, 1, 2, 9, 8, 4, 3, 1]size = len(arr)k = 4ans = calculateMaxSumLength(arr, size, k)print("Max Length ::", ans) # Contributed by Ravi Teja Kuchipudi
<script> // JavaScript program to find Maximum sum of lengths of// non-overlapping subarrays with k as the max element. // Below function calculates the Maximum sum of lengths of// non-overlapping subarrays with k as the max element.function calculateMaxSumLength(arr, n, k){    // maximun sum of lengths    let ans = 0;         // number of elements in current subarray    let count = 0;         // falg variable for checking if k is present    // in current subarray or not    let flag = 0;    for (let i = 0; i < n; i++)    {             // increment the count if element in arr is less        // than k        if (arr[i] < k) {            count++;        }                 // if the element is equals to k        else if (arr[i] == k) {            if (flag == 0) {                count++;                flag = 1;            }                         // if flag is 1, we can say k is already present            // in that subarray. So, add the value of count            // to ans variable and make the count value as 1            // because we found the k            else {                ans += count;                count = 1;            }        }                 // if element in arr is grater than k        else        {                     // if k is present in the subarray            // add the value of count to ans variable            if (flag == 1)            {                ans += count;            }                         // assign value of count and flag variables as 0            count = 0;            flag = 0;        }    }         // Check again if k value is found in subarray    // if k is found, return sum of values of variables ans    // and count if k is not found, return value of variable    // ans.    if (flag == 1) {        return ans + count;    }    return ans;} // driver programlet arr = [ 4, 5, 7, 1, 2, 9, 8, 4, 3, 1 ];let size = arr.length;let k = 4;let ans = calculateMaxSumLength(arr, size, k);document.write("Max Length :: ",ans,"</br>"); // This code is contributed by shinjanpatra </script>
Max Length :: 4
Time Complexity : O(n) 
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