A subarray is called alternating if any two consecutive numbers in it have opposite signs (i.e. one of them should be negative, whereas the other should be positive).
Given an array of n integers. For each index i, we need to find the length if the longest alternating subarray starting at i.
Examples:
Input : a[] = {1, -5, 1, -5}
Output : For index 0, {1, -5, 1, -5} = 4
index 1, {-5, 1, -5} = 3
index 2, {1, -5} = 2
index 3, {-5} = 1.
Input :a[] = {-5, -1, -1, 2, -2, -3}
Output : index 0 = 1,
index 1 = 1,
index 2 = 3,
index 3 = 2,
index 4 = 1,
index 5 = 1,
A Naive approach is to use two loops in which we traverse the whole array starting from every index i (0 to n-1) and calculate the length of the alternating subarray.
Time Complexity: O(n2).
Efficient Approach:
Observe that when a[i] and a[i+1] have opposite signs, count[i] will be 1 more than count[i+1]. Otherwise when they have same sign count[i] will be 1.
We use Dynamic Programming here.
C++
// CPP program to find longest alternating // subarray starting from every index. #include <bits/stdc++.h> using namespace std; void longestAlternating(int arr[], int n) { int count[n]; // Fill count[] from end. count[n - 1] = 1; for (int i = n - 2; i >= 0; i--) { if (arr[i] * arr[i + 1] < 0) count[i] = count[i + 1] + 1; else count[i] = 1; } // Print result for (int i = 0; i < n; i++) cout << count[i] << " "; } // Driver code int main() { int a[] = { -5, -1, -1, 2, -2, -3 }; int n = sizeof(a) / sizeof(a[0]); longestAlternating(a, n); return 0; } |
Java
// Java program to find longest alternating // subarray starting from every index. import java.util.*; class Longest{ public static void longestAlternating(int arr[], int n) { int[] count = new int[n]; // Fill count[] from end. count[n - 1] = 1; for (int i = n - 2; i >= 0; i--) { if (arr[i] * arr[i + 1] < 0) count[i] = count[i + 1] + 1; else count[i] = 1; } // Print result for (int i = 0; i < n; i++) System.out.print(count[i] + " "); } // driver program public static void main(String[] args) { int a[] = { -5, -1, -1, 2, -2, -3 }; int n = 6; longestAlternating(a, n); } } // This code is contributed by rishabh_jain |
Python3
# Python3 program to find longest alternating # subarray starting from every index. def longestAlternating(arr, n) : count = [None] * n # Fill count[] from end. count[n - 1] = 1 i = n - 2 while i >= 0 : if (arr[i] * arr[i + 1] < 0) : count[i] = count[i + 1] + 1 else : count[i] = 1; i = i - 1 i = 0 # Print result while i < n : print (count[i], end = " ") i = i + 1 # Driver Code a = [ -5, -1, -1, 2, -2, -3 ] n = len(a) longestAlternating(a, n); # This code is contributed by rishabh_jain |
C#
//C# program to find longest alternating // subarray starting from every index. using System; class Longest { public static void longestAlternating(int []arr, int n) { int[] count = new int[n]; // Fill count[] from end. count[n - 1] = 1; for (int i = n - 2; i >= 0; i--) { if (arr[i] * arr[i + 1] < 0) count[i] = count[i + 1] + 1; else count[i] = 1; } // Print result for (int i = 0; i < n; i++) Console.Write(count[i] + " "); } // Driver program public static void Main() { int []a = { -5, -1, -1, 2, -2, -3 }; int n = 6; longestAlternating(a, n); } } // This code is contributed by vt_m |
PHP
<?php // PHP program to find longest alternating // subarray starting from every index. function longestAlternating( $arr, $n) { $count = array(); // Fill count[] from end. $count[$n - 1] = 1; for ( $i = $n - 2; $i >= 0; $i--) { if ($arr[$i] * $arr[$i + 1] < 0) $count[$i] = $count[$i + 1] + 1; else $count[$i] = 1; } // Print result for ( $i = 0; $i < $n; $i++) echo $count[$i] , " "; } // Driver code $a = array( -5, -1, -1, 2, -2, -3 ); $n =count($a); longestAlternating($a, $n); // This code is contributed by anuj_67. ?> |
Output:
1 1 3 2 1 1
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Improved By : vt_m
