Given an array containing n numbers. The problem is to find the maximum sum bitonic subarray. A bitonic subarray is a subarray in which elements are first increasing and then decreasing. A strictly increasing or strictly decreasing subarray is also considered a bitonic subarray. Time Complexity of O(n) is required.
Examples:
Input : arr[] = {5, 3, 9, 2, 7, 6, 4}
Output : 19
The subarray is {2, 7, 6, 4}.
Input : arr[] = {9, 12, 14, 8, 6, 5, 10, 20}
Output : 54
Approach: The problem is closely related to the Maximum Sum Bitonic Subsequence. We create two arrays msis[] and msds[]. msis[i] stores the sum of Increasing subarray ending with arr[i]. msds[i] stores the sum of Decreasing subarray starting from arr[i]. Now, maximum sum bitonic subarray is calculated as max(msis[i]+msds[i]-arr[i]) for each index i of the array.
Implementation:
C++
// C++ implementation to find the// maximum sum bitonic subarray#include <bits/stdc++.h>using namespace std;// function to find the maximum sum // bitonic subarrayint maxSumBitonicSubArr(int arr[], int n){ // 'msis[]' to store the maximum sum increasing subarray // up to each index of 'arr' from the beginning // 'msds[]' to store the maximum sum decreasing subarray // from each index of 'arr' up to the end int msis[n], msds[n]; // to store the maximum sum // bitonic subarray int max_sum = INT_MIN; // building up the maximum sum increasing subarray // for each array index msis[0] = arr[0]; for (int i=1; i<n; i++) if (arr[i] > arr[i-1]) msis[i] = msis[i-1] + arr[i]; else msis[i] = arr[i]; // building up the maximum sum decreasing subarray // for each array index msds[n-1] = arr[n-1]; for (int i=n-2; i>=0; i--) if (arr[i] > arr[i+1]) msds[i] = msds[i+1] + arr[i]; else msds[i] = arr[i]; // for each array index, calculating the maximum sum // of bitonic subarray of which it is a part of for (int i=0; i<n; i++) // if true , then update 'max' bitonic // subarray sum if (max_sum < (msis[i] + msds[i] - arr[i])) max_sum = msis[i] + msds[i] - arr[i]; // required maximum sum return max_sum;}// Driver program to test aboveint main(){ int arr[] = {5, 3, 9, 2, 7, 6, 4}; int n = sizeof(arr) / sizeof(arr[0]); cout << "Maximum Sum = " << maxSumBitonicSubArr(arr, n); return 0; } |
Java
// Java implementation to // find the maximum sum// bitonic subarrayclass GFG{ // function to find the maximum // sum bitonic subarray static int maxSumBitonicSubArr(int arr[], int n) { // 'msis[]' to store the maximum // sum increasing subarray up to // each index of 'arr' from the // beginning 'msds[]' to store // the maximum sum decreasing // subarray from each index of // 'arr' up to the end int []msis = new int[n]; int []msds = new int[n]; // to store the maximum // sum bitonic subarray int max_sum = Integer.MIN_VALUE; // building up the maximum // sum increasing subarray // for each array index msis[0] = arr[0]; for (int i = 1; i < n; i++) if (arr[i] > arr[i - 1]) msis[i] = msis[i - 1] + arr[i]; else msis[i] = arr[i]; // building up the maximum // sum decreasing subarray // for each array index msds[n - 1] = arr[n - 1]; for (int i = n - 2; i >= 0; i--) if (arr[i] > arr[i + 1]) msds[i] = msds[i + 1] + arr[i]; else msds[i] = arr[i]; // for each array index, // calculating the maximum // sum of bitonic subarray // of which it is a part of for (int i = 0; i < n; i++) // if true , then update // 'max' bitonic subarray sum if (max_sum < (msis[i] + msds[i] - arr[i])) max_sum = msis[i] + msds[i] - arr[i]; // required maximum sum return max_sum; } // Driver Code public static void main(String[] args) { int arr[] = {5, 3, 9, 2, 7, 6, 4}; int n = arr.length; System.out.println( "Maximum Sum = " + maxSumBitonicSubArr(arr, n)); }}// This code is contributed// by ChitraNayal |
Python 3
# Python 3 implementation # to find the maximum sum# bitonic subarray# function to find the # maximum sum bitonic subarraydef maxSumBitonicSubArr(arr, n): # 'msis[]' to store the maximum # sum increasing subarray up to # each index of 'arr' from the # beginning 'msds[]' to store # the maximum sum decreasing # subarray from each index of # 'arr' up to the end msis = [None] * n msds = [None] * n # to store the maximum # sum bitonic subarray max_sum = 0 # building up the maximum # sum increasing subarray # for each array index msis[0] = arr[0] for i in range(1, n): if (arr[i] > arr[i - 1]): msis[i] = msis[i - 1] + arr[i] else: msis[i] = arr[i] # building up the maximum # sum decreasing subarray # for each array index msds[n - 1] = arr[n - 1] for i in range(n - 2, -1, -1): if (arr[i] > arr[i + 1]): msds[i] = msds[i + 1] + arr[i] else: msds[i] = arr[i] # for each array index, # calculating the maximum # sum of bitonic subarray # of which it is a part of for i in range(n): # if true , then update # 'max' bitonic subarray sum if (max_sum < (msis[i] + msds[i] - arr[i])): max_sum = (msis[i] + msds[i] - arr[i]) # required maximum sum return max_sum# Driver Codearr = [5, 3, 9, 2, 7, 6, 4];n = len(arr)print("Maximum Sum = "+ str(maxSumBitonicSubArr(arr, n))) # This code is contributed# by ChitraNayal |
C#
// C# implementation to find // the maximum sum bitonic subarrayusing System;class GFG{ // function to find the maximum // sum bitonic subarray static int maxSumBitonicSubArr(int[] arr, int n) { // 'msis[]' to store the maximum // sum increasing subarray up to // each index of 'arr' from the // beginning 'msds[]' to store // the maximum sum decreasing // subarray from each index of // 'arr' up to the end int []msis = new int[n]; int []msds = new int[n]; // to store the maximum // sum bitonic subarray int max_sum = int.MinValue; // building up the maximum // sum increasing subarray // for each array index msis[0] = arr[0]; for (int i = 1; i < n; i++) if (arr[i] > arr[i - 1]) msis[i] = msis[i - 1] + arr[i]; else msis[i] = arr[i]; // building up the maximum // sum decreasing subarray // for each array index msds[n - 1] = arr[n - 1]; for (int i = n - 2; i >= 0; i--) if (arr[i] > arr[i + 1]) msds[i] = msds[i + 1] + arr[i]; else msds[i] = arr[i]; // for each array index, calculating // the maximum sum of bitonic subarray // of which it is a part of for (int i = 0; i < n; i++) // if true , then update // 'max' bitonic subarray sum if (max_sum < (msis[i] + msds[i] - arr[i])) max_sum = msis[i] + msds[i] - arr[i]; // required maximum sum return max_sum; } // Driver Code public static void Main() { int[] arr = {5, 3, 9, 2, 7, 6, 4}; int n = arr.Length; Console.Write("Maximum Sum = " + maxSumBitonicSubArr(arr, n)); }}// This code is contributed// by ChitraNayal |
PHP
<?php// PHP implementation to find the// maximum sum bitonic subarray// function to find the maximum sum // bitonic subarrayfunction maxSumBitonicSubArr($arr, $n){ // 'msis[]' to store the maximum // sum increasing subarray up to // each index of 'arr' from the // beginning 'msds[]' to store // the maximum sum decreasing // subarray from each index of // 'arr' up to the end $msis = array(); $msds = array(); // to store the maximum // sum bitonic subarray $max_sum = PHP_INT_MIN; // building up the maximum // sum increasing subarray // for each array index $msis[0] = $arr[0]; for ($i = 1; $i < $n; $i++) if ($arr[$i] > $arr[$i - 1]) $msis[$i] = $msis[$i - 1] + $arr[$i]; else $msis[$i] = $arr[$i]; // building up the maximum // sum decreasing subarray // for each array index $msds[$n - 1] = $arr[$n - 1]; for ($i = $n - 2; $i >= 0; $i--) if ($arr[$i] > $arr[$i + 1]) $msds[$i] = $msds[$i + 1] + $arr[$i]; else $msds[$i] = $arr[$i]; // for each array index, // calculating the maximum sum // of bitonic subarray of which // it is a part of for ($i = 0; $i < $n; $i++) // if true , then update // 'max' bitonic subarray sum if ($max_sum < ($msis[$i] + $msds[$i] - $arr[$i])) $max_sum = $msis[$i] + $msds[$i] - $arr[$i]; // required maximum sum return $max_sum;}// Driver Code$arr = array(5, 3, 9, 2, 7, 6, 4);$n = sizeof($arr);echo "Maximum Sum = ", maxSumBitonicSubArr($arr, $n);// This code is contributed by ajit?> |
Javascript
<script> // Javascript implementation to find // the maximum sum bitonic subarray // function to find the maximum // sum bitonic subarray function maxSumBitonicSubArr(arr, n) { // 'msis[]' to store the maximum // sum increasing subarray up to // each index of 'arr' from the // beginning 'msds[]' to store // the maximum sum decreasing // subarray from each index of // 'arr' up to the end let msis = new Array(n); msis.fill(0); let msds = new Array(n); msds.fill(0); // to store the maximum // sum bitonic subarray let max_sum = Number.MIN_VALUE; // building up the maximum // sum increasing subarray // for each array index msis[0] = arr[0]; for (let i = 1; i < n; i++) if (arr[i] > arr[i - 1]) msis[i] = msis[i - 1] + arr[i]; else msis[i] = arr[i]; // building up the maximum // sum decreasing subarray // for each array index msds[n - 1] = arr[n - 1]; for (let i = n - 2; i >= 0; i--) if (arr[i] > arr[i + 1]) msds[i] = msds[i + 1] + arr[i]; else msds[i] = arr[i]; // for each array index, calculating // the maximum sum of bitonic subarray // of which it is a part of for (let i = 0; i < n; i++) // if true , then update // 'max' bitonic subarray sum if (max_sum < (msis[i] + msds[i] - arr[i])) max_sum = msis[i] + msds[i] - arr[i]; // required maximum sum return max_sum; } let arr = [5, 3, 9, 2, 7, 6, 4]; let n = arr.length; document.write("Maximum Sum = " + maxSumBitonicSubArr(arr, n)); </script> |
Output
Maximum Sum = 19
Time Complexity: O(n)
Auxiliary Space: O(n)
Space optimized solution: It can be solved with constant memory. Indeed, since we are looking for contiguous subarrays, we can separate the initial array into bitonic chunks and compare their sums.
Implementation:
C++
// C++ implementation to find the// maximum sum bitonic subarray#include <bits/stdc++.h>using namespace std;// Function to find the maximum sum bitonic// subarray.int maxSumBitonicSubArr(int arr[], int n){ // to store the maximum sum // bitonic subarray int max_sum = INT_MIN; int i = 0; while (i < n) { // Find the longest increasing subarray // starting at i. int j = i; while (j+1 < n && arr[j] < arr[j+1]) j++; // Now we know that a[i..j] is an // increasing subarray. Remove non- // positive elements from the left // side as much as possible. while (i < j && arr[i] <= 0) i++; // Find the longest decreasing subarray // starting at j. int k = j; while (k+1 < n && arr[k] > arr[k+1]) k++; // Now we know that a[j..k] is a // decreasing subarray. Remove non- // positive elements from the right // side as much as possible. // last is needed to keep the last // seen element. int last = k; while (k > j && arr[k] <= 0) k--; // Compute the max sum of the // increasing part. int sum_inc = accumulate(arr+i, arr+j+1, 0); // Compute the max sum of the // decreasing part. int sum_dec = accumulate(arr+j, arr+k+1, 0); // The overall max sum is the sum of // both parts minus the peak element, // because it was counted twice. int sum_all = sum_inc + sum_dec - arr[j]; max_sum = max({max_sum, sum_inc, sum_dec, sum_all}); // If the next element is equal to the // current, i.e. arr[i+1] == arr[i], // last == i. // To ensure the algorithm has progress, // get the max of last and i+1. i = max(last, i+1); } // required maximum sum return max_sum;}// Driver program to test aboveint main(){ // The example from the article, the // answer is 19. int arr[] = {5, 3, 9, 2, 7, 6, 4}; int n = sizeof(arr) / sizeof(arr[0]); cout << "Maximum Sum = " << maxSumBitonicSubArr(arr, n) << endl; // Always increasing, the answer is 15. int arr2[] = {1, 2, 3, 4, 5}; int n2 = sizeof(arr2) / sizeof(arr2[0]); cout << "Maximum Sum = " << maxSumBitonicSubArr(arr2, n2) << endl; // Always decreasing, the answer is 15. int arr3[] = {5, 4, 3, 2, 1}; int n3 = sizeof(arr3) / sizeof(arr3[0]); cout << "Maximum Sum = " << maxSumBitonicSubArr(arr3, n3) << endl; // All are equal, the answer is 5. int arr4[] = {5, 5, 5, 5}; int n4 = sizeof(arr4) / sizeof(arr4[0]); cout << "Maximum Sum = " << maxSumBitonicSubArr(arr4, n4) << endl; // The whole array is bitonic, but the answer is 7. int arr5[] = {-1, 0, 1, 2, 3, 1, 0, -1, -10}; int n5 = sizeof(arr5) / sizeof(arr5[0]); cout << "Maximum Sum = " << maxSumBitonicSubArr(arr5, n5) << endl; // The answer is 4 (the tail). int arr6[] = {-1, 0, 1, 2, 0, -1, -2, 0, 1, 3}; int n6 = sizeof(arr6) / sizeof(arr6[0]); cout << "Maximum Sum = " << maxSumBitonicSubArr(arr6, n6) << endl; return 0;} |
Java
// Java implementation to find the // maximum sum bitonic subarray import java.util.*; class GFG{ static int find_partial_sum(int arr[], int start, int end){ int sum = 0; for(int i = start; i < end; i++) sum += arr[i]; return sum;}// Function to find the maximum sum bitonic // subarray. static int maxSumBitonicSubArr(int arr[], int n) { // To store the maximum sum // bitonic subarray int max_sum = -1000000; int i = 0; while (i < n) { // Find the longest increasing // subarray starting at i. int j = i; while (j + 1 < n && arr[j] < arr[j + 1]) j++; // Now we know that a[i..j] is an // increasing subarray. Remove non- // positive elements from the left // side as much as possible. while (i < j && arr[i] <= 0) i++; // Find the longest decreasing subarray // starting at j. int k = j; while (k + 1 < n && arr[k] > arr[k + 1]) k++; // Now we know that a[j..k] is a // decreasing subarray. Remove non- // positive elements from the right // side as much as possible. // last is needed to keep the last // seen element. int last = k; while (k > j && arr[k] <= 0) k--; // Compute the max sum of the // increasing part. int sum_inc = find_partial_sum(arr, i, j + 1); // Compute the max sum of the // decreasing part. int sum_dec = find_partial_sum(arr, j, k + 1); // The overall max sum is the sum of // both parts minus the peak element, // because it was counted twice. int sum_all = sum_inc + sum_dec - arr[j]; max_sum = Math.max(Math.max(max_sum, sum_inc), Math.max(sum_dec, sum_all)); // If the next element is equal to the // current, i.e. arr[i+1] == arr[i], // last == i. // To ensure the algorithm has progress, // get the max of last and i+1. i = Math.max(last, i + 1); } // Required maximum sum return max_sum; } // Driver code public static void main(String args[]) { // The example from the article, the // answer is 19. int arr[] = { 5, 3, 9, 2, 7, 6, 4 }; int n = arr.length; System.out.println("Maximum sum = " + maxSumBitonicSubArr(arr, n)); // Always increasing, the answer is 15. int arr2[] = { 1, 2, 3, 4, 5 }; int n2 = arr2.length; System.out.println("Maximum sum = " + maxSumBitonicSubArr(arr2, n2)); // Always decreasing, the answer is 15. int arr3[] = { 5, 4, 3, 2, 1 }; int n3 = arr3.length; System.out.println("Maximum sum = " + maxSumBitonicSubArr(arr3, n3)); // All are equal, the answer is 5. int arr4[] = { 5, 5, 5, 5 }; int n4 = arr4.length; System.out.println("Maximum sum = " + maxSumBitonicSubArr(arr4, n4)); // The whole array is bitonic, // but the answer is 7. int arr5[] = { -1, 0, 1, 2, 3, 1, 0, -1, -10 }; int n5 = arr5.length; System.out.println("Maximum sum = " + maxSumBitonicSubArr(arr5, n5)); // The answer is 4 (the tail). int arr6[] = { -1, 0, 1, 2, 0, -1, -2, 0, 1, 3 }; int n6 = arr6.length; System.out.println("Maximum sum = " + maxSumBitonicSubArr(arr6, n6)); }}// This code is contributed by amreshkumar3 |
Python3
# Python3 implementation to find the# maximum sum bitonic subarray# Function to find the maximum sum bitonic# subarray.def maxSumBitonicSubArr(arr, n): # to store the maximum sum # bitonic subarray max_sum = -10**9 i = 0 while (i < n): # Find the longest increasing subarray # starting at i. j = i while (j + 1 < n and arr[j] < arr[j + 1]): j += 1 # Now we know that a[i..j] is an # increasing subarray. Remove non- # positive elements from the left # side as much as possible. while (i < j and arr[i] <= 0): i += 1 # Find the longest decreasing subarray # starting at j. k = j while (k + 1 < n and arr[k] > arr[k + 1]): k += 1 # Now we know that a[j..k] is a # decreasing subarray. Remove non- # positive elements from the right # side as much as possible. # last is needed to keep the last # seen element. last = k while (k > j and arr[k] <= 0): k -= 1 # Compute the max sum of the # increasing part. nn = arr[i:j + 1] sum_inc = sum(nn) # Compute the max sum of the # decreasing part. nn = arr[j:k + 1] sum_dec = sum(nn) # The overall max sum is the sum of # both parts minus the peak element, # because it was counted twice. sum_all = sum_inc + sum_dec - arr[j] max_sum = max([max_sum, sum_inc, sum_dec, sum_all]) # If the next element is equal to the # current, i.e. arr[i+1] == arr[i], # last == i. # To ensure the algorithm has progress, # get the max of last and i+1. i = max(last, i + 1) # required maximum sum return max_sum# Driver Code# The example from the article, the# answer is 19.arr = [5, 3, 9, 2, 7, 6, 4]n = len(arr)print("Maximum Sum = ", maxSumBitonicSubArr(arr, n)) # Always increasing, the answer is 15.arr2 = [1, 2, 3, 4, 5]n2 = len(arr2)print("Maximum Sum = ", maxSumBitonicSubArr(arr2, n2))# Always decreasing, the answer is 15.arr3 = [5, 4, 3, 2, 1]n3 = len(arr3)print("Maximum Sum = ", maxSumBitonicSubArr(arr3, n3))# All are equal, the answer is 5.arr4 = [5, 5, 5, 5]n4 = len(arr4)print("Maximum Sum = ", maxSumBitonicSubArr(arr4, n4))# The whole array is bitonic, # but the answer is 7.arr5 = [-1, 0, 1, 2, 3, 1, 0, -1, -10]n5 = len(arr5)print("Maximum Sum = ", maxSumBitonicSubArr(arr5, n5))# The answer is 4 (the tail).arr6 = [-1, 0, 1, 2, 0, -1, -2, 0, 1, 3]n6 = len(arr6)print("Maximum Sum = ", maxSumBitonicSubArr(arr6, n6))# This code is contributed by Mohit Kumar |
C#
// C# implementation to find the // maximum sum bitonic subarray using System; class GFG{ static int find_partial_sum(int []arr, int start, int end){ int sum = 0; for(int i = start; i < end; i++) sum += arr[i]; return sum;}// Function to find the maximum sum bitonic // subarray. static int maxSumBitonicSubArr(int []arr, int n) { // To store the maximum sum // bitonic subarray int max_sum = -1000000; int i = 0; while (i < n) { // Find the longest increasing subarray // starting at i. int j = i; while (j + 1 < n && arr[j] < arr[j + 1]) j++; // Now we know that a[i..j] is an // increasing subarray. Remove non- // positive elements from the left // side as much as possible. while (i < j && arr[i] <= 0) i++; // Find the longest decreasing subarray // starting at j. int k = j; while (k + 1 < n && arr[k] > arr[k + 1]) k++; // Now we know that a[j..k] is a // decreasing subarray. Remove non- // positive elements from the right // side as much as possible. // last is needed to keep the last // seen element. int last = k; while (k > j && arr[k] <= 0) k--; // Compute the max sum of the // increasing part. int sum_inc = find_partial_sum(arr, i, j + 1); // Compute the max sum of the // decreasing part. int sum_dec = find_partial_sum(arr, j, k + 1); // The overall max sum is the sum of // both parts minus the peak element, // because it was counted twice. int sum_all = sum_inc + sum_dec - arr[j]; max_sum = Math.Max(Math.Max(max_sum, sum_inc), Math.Max(sum_dec, sum_all)); // If the next element is equal to the // current, i.e. arr[i+1] == arr[i], // last == i. // To ensure the algorithm has progress, // get the max of last and i+1. i = Math.Max(last, i + 1); } // Required maximum sum return max_sum; } // Driver code public static void Main() { // The example from the article, the // answer is 19. int []arr = { 5, 3, 9, 2, 7, 6, 4 }; int n = arr.Length; Console.WriteLine("Maximum sum = " + maxSumBitonicSubArr(arr, n)); // Always increasing, the answer is 15. int []arr2 = { 1, 2, 3, 4, 5 }; int n2 = arr2.Length; Console.WriteLine("Maximum sum = " + maxSumBitonicSubArr(arr2, n2)); // Always decreasing, the answer is 15. int []arr3 = { 5, 4, 3, 2, 1 }; int n3 = arr3.Length; Console.WriteLine("Maximum sum = " + maxSumBitonicSubArr(arr3, n3)); // All are equal, the answer is 5. int []arr4 = { 5, 5, 5, 5 }; int n4 = arr4.Length; Console.WriteLine("Maximum sum = " + maxSumBitonicSubArr(arr4, n4)); // The whole array is bitonic, // but the answer is 7. int []arr5 = { -1, 0, 1, 2, 3, 1, 0, -1, -10 }; int n5 = arr5.Length; Console.WriteLine("Maximum sum = " + maxSumBitonicSubArr(arr5, n5)); // The answer is 4 (the tail). int []arr6 = { -1, 0, 1, 2, 0, -1, -2, 0, 1, 3}; int n6 = arr6.Length; Console.WriteLine("Maximum sum = " + maxSumBitonicSubArr(arr6, n6)); }}// This code is contributed by amreshkumar3 |
Javascript
<script>// Javascript implementation to find the// maximum sum bitonic subarray function find_partial_sum(arr,start,end) { let sum = 0; for(let i = start; i < end; i++) sum += arr[i]; return sum; }// Function to find the maximum sum bitonic// subarray.function maxSumBitonicSubArr(arr,n){ // To store the maximum sum // bitonic subarray let max_sum = -1000000; let i = 0; while (i < n) { // Find the longest increasing // subarray starting at i. let j = i; while (j + 1 < n && arr[j] < arr[j + 1]) j++; // Now we know that a[i..j] is an // increasing subarray. Remove non- // positive elements from the left // side as much as possible. while (i < j && arr[i] <= 0) i++; // Find the longest decreasing subarray // starting at j. let k = j; while (k + 1 < n && arr[k] > arr[k + 1]) k++; // Now we know that a[j..k] is a // decreasing subarray. Remove non- // positive elements from the right // side as much as possible. // last is needed to keep the last // seen element. let last = k; while (k > j && arr[k] <= 0) k--; // Compute the max sum of the // increasing part. let sum_inc = find_partial_sum(arr, i, j + 1); // Compute the max sum of the // decreasing part. let sum_dec = find_partial_sum(arr, j, k + 1); // The overall max sum is the sum of // both parts minus the peak element, // because it was counted twice. let sum_all = sum_inc + sum_dec - arr[j]; max_sum = Math.max(Math.max(max_sum, sum_inc), Math.max(sum_dec, sum_all)); // If the next element is equal to the // current, i.e. arr[i+1] == arr[i], // last == i. // To ensure the algorithm has progress, // get the max of last and i+1. i = Math.max(last, i + 1); } // Required maximum sum return max_sum;}// The example from the article, the// answer is 19.let arr = [ 5, 3, 9, 2, 7, 6, 4 ];let n = arr.length;document.write("Maximum sum = " + maxSumBitonicSubArr(arr, n)+"<br>");// Always increasing, the answer is 15.let arr2 = [ 1, 2, 3, 4, 5 ];let n2 = arr2.length;document.write("Maximum sum = " + maxSumBitonicSubArr(arr2, n2)+"<br>");// Always decreasing, the answer is 15.let arr3 = [ 5, 4, 3, 2, 1 ];let n3 = arr3.length;document.write("Maximum sum = " + maxSumBitonicSubArr(arr3, n3)+"<br>");// All are equal, the answer is 5.let arr4 = [ 5, 5, 5, 5 ];let n4 = arr4.length;document.write("Maximum sum = " + maxSumBitonicSubArr(arr4, n4)+"<br>");// The whole array is bitonic,// but the answer is 7.let arr5 = [ -1, 0, 1, 2, 3, 1, 0, -1, -10 ];let n5 = arr5.length;document.write("Maximum sum = " + maxSumBitonicSubArr(arr5, n5)+"<br>");// The answer is 4 (the tail).let arr6 = [ -1, 0, 1, 2, 0, -1, -2, 0, 1, 3 ];let n6 = arr6.length;document.write("Maximum sum = " + maxSumBitonicSubArr(arr6, n6)+"<br>");// This code is contributed by patel2127</script> |
Output
Maximum Sum = 19 Maximum Sum = 15 Maximum Sum = 15 Maximum Sum = 5 Maximum Sum = 7 Maximum Sum = 4
Time Complexity: The algorithm uses a single while loop to iterate through the array and nested loops to find the increasing and decreasing subarrays, so the time complexity is O(n).
Space Complexity: The algorithm uses constant extra space, so the space complexity is O(1).
Thanks to Andrey Khayrutdinov for suggesting this solution.
This article is contributed by Ayush Jauhari. If you like neveropen 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 neveropen main page and help other Geeks.
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
