Given an array arr[] and an integer X, the task is to print the longest subarray such that the sum of its elements isn’t divisible by X. If no such subarray exists, print “-1”.
Note: If more than one subarray exists with the given property, print any one of them.
Examples:
Input: arr[] = {1, 2, 3} X = 3
Output: 2 3
Explanation:
The subarray {2, 3} has a sum of elements 5, which isn’t divisible by 3.
Input: arr[] = {2, 6} X = 2
Output: -1
Explanation:
All possible subarrays {1}, {2}, {1, 2} have an even sum.
Therefore, the answer is -1.
Naive Approach: The simplest approach to solve the problem is to generate all possible subarrays and keep calculating its sum. If any subarray is found to have sum not divisible by X, compare the length with maximum length obtained(maxm) and update the maxm accordingly and update the starting index and ending index of the subarray. Finally, print the subarray having the stored starting and ending indices. If there is no such subarray then print “-1”.
Time Complexity: O(N2)
Auxiliary Space: O(1)
Efficient Approach: To optimize the above approach we will find the prefix and suffix array sum. Follow the steps below:
- Generate the prefix sum array and suffix sum array.
- Iterate from [0, N – 1] using Two Pointers and choose the prefix and suffix sum of the element at each index which is not divisible by X. Store the starting index and ending index of the subarray.
- After completing the above steps, if there exist a subarray with sum not divisible by X, then print the subarray having the stored starting and ending indices.
- If there is no such subarray then print “-1”.
Below is the implementation of the above approach:
C++
#include <iostream> // C++ Program to implement // the above approach #include <bits/stdc++.h> using namespace std; // Function to print the longest // subarray with sum of elements // not divisible by X void max_length( int N, int x, vector< int >& v) { int i, a; // Pref[] stores the prefix sum // Suff[] stores the suffix sum vector< int > preff, suff; int ct = 0; for (i = 0; i < N; i++) { a = v[i]; // If array element is // divisibile by x if (a % x == 0) { // Increase count ct += 1; } } // If all the array elements // are divisible by x if (ct == N) { // No subarray possible cout << -1 << endl; return ; } // Reverse v to calculate the // suffix sum reverse(v.begin(), v.end()); suff.push_back(v[0]); // Calculate the suffix sum for (i = 1; i < N; i++) { suff.push_back(v[i] + suff[i - 1]); } // Reverse to original form reverse(v.begin(), v.end()); // Reverse the suffix sum array reverse(suff.begin(), suff.end()); preff.push_back(v[0]); // Calculate the prefix sum for (i = 1; i < N; i++) { preff.push_back(v[i] + preff[i - 1]); } int ans = 0; // Stores the starting index // of required subarray int lp = 0; // Stores the ending index // of required subarray int rp = N - 1; for (i = 0; i < N; i++) { // If suffix sum till i-th // index is not divisible by x if (suff[i] % x != 0 && (ans < (N - 1))) { lp = i; rp = N - 1; // Update the answer ans = max(ans, N - i); } // If prefix sum till i-th // index is not divisible by x if (preff[i] % x != 0 && (ans < (i + 1))) { lp = 0; rp = i; // Update the answer ans = max(ans, i + 1); } } // Print the longest subarray for (i = lp; i <= rp; i++) { cout << v[i] << " " ; } } // Driver Code int main() { int x = 3; vector< int > v = { 1, 3, 2, 6 }; int N = v.size(); max_length(N, x, v); return 0; } |
Java
// Java program to implement // the above approach import java.util.*; class GFG{ // Function to print the longest // subarray with sum of elements // not divisible by X static void max_length( int N, int x, int []v) { int i, a; // Pref[] stores the prefix sum // Suff[] stores the suffix sum List<Integer> preff = new Vector<Integer>(); List<Integer> suff = new Vector<Integer>(); int ct = 0 ; for (i = 0 ; i < N; i++) { a = v[i]; // If array element is // divisibile by x if (a % x == 0 ) { // Increase count ct += 1 ; } } // If all the array elements // are divisible by x if (ct == N) { // No subarray possible System.out.print(- 1 + "\n" ); return ; } // Reverse v to calculate the // suffix sum v = reverse(v); suff.add(v[ 0 ]); // Calculate the suffix sum for (i = 1 ; i < N; i++) { suff.add(v[i] + suff.get(i - 1 )); } // Reverse to original form v = reverse(v); // Reverse the suffix sum array Collections.reverse(suff); preff.add(v[ 0 ]); // Calculate the prefix sum for (i = 1 ; i < N; i++) { preff.add(v[i] + preff.get(i - 1 )); } int ans = 0 ; // Stores the starting index // of required subarray int lp = 0 ; // Stores the ending index // of required subarray int rp = N - 1 ; for (i = 0 ; i < N; i++) { // If suffix sum till i-th // index is not divisible by x if (suff.get(i) % x != 0 && (ans < (N - 1 ))) { lp = i; rp = N - 1 ; // Update the answer ans = Math.max(ans, N - i); } // If prefix sum till i-th // index is not divisible by x if (preff.get(i) % x != 0 && (ans < (i + 1 ))) { lp = 0 ; rp = i; // Update the answer ans = Math.max(ans, i + 1 ); } } // Print the longest subarray for (i = lp; i <= rp; i++) { System.out.print(v[i] + " " ); } } static int [] reverse( int a[]) { int i, n = a.length, t; for (i = 0 ; i < n / 2 ; i++) { t = a[i]; a[i] = a[n - i - 1 ]; a[n - i - 1 ] = t; } return a; } // Driver Code public static void main(String[] args) { int x = 3 ; int []v = { 1 , 3 , 2 , 6 }; int N = v.length; max_length(N, x, v); } } // This code is contributed by PrinciRaj1992 |
Python3
# Python3 program to implement # the above approach # Function to print the longest # subarray with sum of elements # not divisible by X def max_length(N, x, v): # Pref[] stores the prefix sum # Suff[] stores the suffix sum preff, suff = [], [] ct = 0 for i in range (N): a = v[i] # If array element is # divisibile by x if a % x = = 0 : # Increase count ct + = 1 # If all the array elements # are divisible by x if ct = = N: # No subarray possible print ( - 1 ) return # Reverse v to calculate the # suffix sum v.reverse() suff.append(v[ 0 ]) # Calculate the suffix sum for i in range ( 1 , N): suff.append(v[i] + suff[i - 1 ]) # Reverse to original form v.reverse() # Reverse the suffix sum array suff.reverse() preff.append(v[ 0 ]) # Calculate the prefix sum for i in range ( 1 , N): preff.append(v[i] + preff[i - 1 ]) ans = 0 # Stores the starting index # of required subarray lp = 0 # Stores the ending index # of required subarray rp = N - 1 for i in range (N): # If suffix sum till i-th # index is not divisible by x if suff[i] % x ! = 0 and ans < N - 1 : lp = i rp = N - 1 # Update the answer ans = max (ans, N - i) # If prefix sum till i-th # index is not divisible by x if preff[i] % x ! = 0 and ans < i + 1 : lp = 0 rp = i # Update the answer ans = max (ans, i + 1 ) # Print the longest subarray for i in range (lp, rp + 1 ): print (v[i], end = " " ) # Driver code x = 3 v = [ 1 , 3 , 2 , 6 ] N = len (v) max_length(N, x, v) # This code is contributed by Stuti Pathak |
C#
// C# program to implement // the above approach using System; using System.Collections.Generic; class GFG{ // Function to print the longest // subarray with sum of elements // not divisible by X static void max_length( int N, int x, int []v) { int i, a; // Pref[] stores the prefix sum // Suff[] stores the suffix sum List< int > preff = new List< int >(); List< int > suff = new List< int >(); int ct = 0; for (i = 0; i < N; i++) { a = v[i]; // If array element is // divisibile by x if (a % x == 0) { // Increase count ct += 1; } } // If all the array elements // are divisible by x if (ct == N) { // No subarray possible Console.Write(-1 + "\n" ); return ; } // Reverse v to calculate the // suffix sum v = reverse(v); suff.Add(v[0]); // Calculate the suffix sum for (i = 1; i < N; i++) { suff.Add(v[i] + suff[i - 1]); } // Reverse to original form v = reverse(v); // Reverse the suffix sum array suff.Reverse(); preff.Add(v[0]); // Calculate the prefix sum for (i = 1; i < N; i++) { preff.Add(v[i] + preff[i - 1]); } int ans = 0; // Stores the starting index // of required subarray int lp = 0; // Stores the ending index // of required subarray int rp = N - 1; for (i = 0; i < N; i++) { // If suffix sum till i-th // index is not divisible by x if (suff[i] % x != 0 && (ans < (N - 1))) { lp = i; rp = N - 1; // Update the answer ans = Math.Max(ans, N - i); } // If prefix sum till i-th // index is not divisible by x if (preff[i] % x != 0 && (ans < (i + 1))) { lp = 0; rp = i; // Update the answer ans = Math.Max(ans, i + 1); } } // Print the longest subarray for (i = lp; i <= rp; i++) { Console.Write(v[i] + " " ); } } static int [] reverse( int []a) { int i, n = a.Length, t; for (i = 0; i < n / 2; i++) { t = a[i]; a[i] = a[n - i - 1]; a[n - i - 1] = t; } return a; } // Driver Code public static void Main(String[] args) { int x = 3; int []v = { 1, 3, 2, 6 }; int N = v.Length; max_length(N, x, v); } } // This code is contributed by PrinciRaj1992 |
3 2 6
Time Complexity: O(N)
Auxiliary Space: O(N)
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