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Given an array A[0 ... n-1] containing n positive integers, a subarray A[i ... j] is bitonic if there is a k with i <= k <= j such that A[i] <= A[i + 1] ... <= A[k] >= A[k + 1] >= .. A[j - 1] > = A[j]. Write a function that takes an array as argument and returns the length of the maximum length bitonic subarray.
Expected time complexity of the solution is O(n)
Simple Examples
1) A[] = {12, 4, 78, 90, 45, 23}, the maximum length bitonic subarray is {4, 78, 90, 45, 23} which is of length 5.
2) A[] = {20, 4, 1, 2, 3, 4, 2, 10}, the maximum length bitonic subarray is {1, 2, 3, 4, 2} which is of length 5.
Extreme Examples
1) A[] = {10}, the single element is bitnoic, so output is 1.
2) A[] = {10, 20, 30, 40}, the complete array itself is bitonic, so output is 4.
3) A[] = {40, 30, 20, 10}, the complete array itself is bitonic, so output is 4.
Solution
Let us consider the array {12, 4, 78, 90, 45, 23} to understand the soultion.
1) Construct an auxiliary array inc[] from left to right such that inc[i] contains length of the nondecreaing subarray ending at arr[i].
For for A[] = {12, 4, 78, 90, 45, 23}, inc[] is {1, 1, 2, 3, 1, 1}
2) Construct another array dec[] from right to left such that dec[i] contains length of nonincreasing subarray starting at arr[i].
For A[] = {12, 4, 78, 90, 45, 23}, dec[] is {2, 1, 1, 3, 2, 1}.
3) Once we have the inc[] and dec[] arrays, all we need to do is find the maximum value of (inc[i] + dec[i] – 1).
For {12, 4, 78, 90, 45, 23}, the max value of (inc[i] + dec[i] – 1) is 5 for i = 3.
Time Complexity: O(n)
Auxiliary Space: O(n)
Given an array A[0 ... n-1] containing n positive integers, a subarray A[i ... j] is bitonic if there is a k with i <= k <= j such that A[i] <= A[i + 1] ... <= A[k] >= A[k + 1] >= .. A[j - 1] > = A[j]. Write a function that takes an array as argument and returns the length of the maximum length bitonic subarray.
Expected time complexity of the solution is O(n)
Simple Examples
1) A[] = {12, 4, 78, 90, 45, 23}, the maximum length bitonic subarray is {4, 78, 90, 45, 23} which is of length 5.
2) A[] = {20, 4, 1, 2, 3, 4, 2, 10}, the maximum length bitonic subarray is {1, 2, 3, 4, 2} which is of length 5.
Extreme Examples
1) A[] = {10}, the single element is bitnoic, so output is 1.
2) A[] = {10, 20, 30, 40}, the complete array itself is bitonic, so output is 4.
3) A[] = {40, 30, 20, 10}, the complete array itself is bitonic, so output is 4.
Solution
Let us consider the array {12, 4, 78, 90, 45, 23} to understand the soultion.
1) Construct an auxiliary array inc[] from left to right such that inc[i] contains length of the nondecreaing subarray ending at arr[i].
For for A[] = {12, 4, 78, 90, 45, 23}, inc[] is {1, 1, 2, 3, 1, 1}
2) Construct another array dec[] from right to left such that dec[i] contains length of nonincreasing subarray starting at arr[i].
For A[] = {12, 4, 78, 90, 45, 23}, dec[] is {2, 1, 1, 3, 2, 1}.
3) Once we have the inc[] and dec[] arrays, all we need to do is find the maximum value of (inc[i] + dec[i] – 1).
For {12, 4, 78, 90, 45, 23}, the max value of (inc[i] + dec[i] – 1) is 5 for i = 3.
#include<stdio.h> #include<stdlib.h> int bitonic( int arr[], int n) { int i; int *inc = new int [n]; int *dec = new int [n]; int max; inc[0] = 1; // The length of increasing sequence ending at first index is 1 dec[n-1] = 1; // The length of increasing sequence starting at first index is 1 // Step 1) Construct increasing sequence array for (i = 1; i < n; i++) { if (arr[i] > arr[i-1]) inc[i] = inc[i-1] + 1; else inc[i] = 1; } // Step 2) Construct decreasing sequence array for (i = n-2; i >= 0; i--) { if (arr[i] > arr[i+1]) dec[i] = dec[i+1] + 1; else dec[i] = 1; } // Step 3) Find the length of maximum length bitonic sequence max = inc[0] + dec[0] - 1; for (i = 1; i < n; i++) { if (inc[i] + dec[i] - 1 > max) { max = inc[i] + dec[i] - 1; } } // free dynamically allocated memory delete [] inc; delete [] dec; return max; } /* Driver program to test above function */ int main() { int arr[] = {12, 4, 78, 90, 45, 23}; int n = sizeof (arr)/ sizeof (arr[0]); printf ( "\n Length of max length Bitnoic Subarray is %d" , bitonic(arr, n)); getchar (); return 0; } |
Auxiliary Space: O(n)
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