Given an array arr of N integers, the task is to calculate the length of longest sequence of consecutive integers that can be formed from the array. It is also given that 0’s in the array can be converted to any value.
Example:
Input: N = 7, A = {0, 6, 5, 10, 3, 0, 11}
Output: 5
Explanation: The maximum consecutive sequence formed can be {3, 4, 5, 6, 7}. As 4, 7 are not present in the array, we can change 2 zeroes to 4 and 7.Input: N = 6, A = {0, 0, 1, 2, 6, 0}
Output: 6
Explanation: The maximum consecutive sequence formed can be {1, 2, 3, 4, 5, 6}
Approach: Given problem can be solved with the help of binary search and prefix sum array:
- Calculate total zeroes in the array and store them in a variable count which indicates the total possible changes that can be made
- Repeating and zero values in the given array are removed so that array contains only unique non-zero values
- Create an auxiliary array and initialize those indices value to 1, whose values are present in the given array
- Auxiliary array is turned into prefix sum array
- Iterate the prefix array and at every iteration perform binary search with lower limit as current index and upper limit as last index of the array
- Let current index be l and the rightmost possible index as r. For each mid = (l + r) / 2, Check if this range[l, mid] is achievable with total allowed changes
- Update l = mid +1 if the above statement is true otherwise r = mid – 1
- Calculate maximum length for all starting values
Below is the implementation of the above approach:
C++
// C++ implementation for the above approach#include <bits/stdc++.h>using namespace std;// Function to calculate maximum// possible consecutive numbers// with changes allowedint maximumConsecutiveNumbers(int arr[], int N){ // Store all non-zero elements // in a new vector and // calculate total zero elements vector<int> v; // Variable to store the // count of zero elements int count = 0; for (int i = 0; i < N; i++) { if (arr[i] == 0) { count++; } else { v.push_back(arr[i]); } } // Sort the array sort(v.begin(), v.end()); // Remove all the duplicates // from the array (v).erase(unique(v.begin(), v.end()), (v).end()); // Variable to store the maximum // value of the sequence int MAXN = 1100000; // Make the prefix array vector<int> pref(MAXN + 1, 0); for (int i = 0; i < v.size(); i++) { pref[v[i]]++; } for (int i = 1; i <= MAXN; i++) { pref[i] += pref[i - 1]; } int mx = 0; // Iterate for each element and // use binary search for (int i = 1; i <= MAXN; i++) { int l = i, r = MAXN; int local_max = 0; while (l <= r) { int mid = (l + r) / 2; // Conversions equal to number // of zeros, can be made upto mid if (pref[mid] - pref[i - 1] + count >= (mid - i + 1)) { l = mid + 1; local_max = max(local_max, mid - i + 1); } else { r = mid - 1; } } mx = max(mx, local_max); } return mx;}// Driver Codeint main(){ int N = 7; int arr[] = { 0, 6, 5, 10, 3, 0, 11 }; cout << maximumConsecutiveNumbers(arr, N);} |
Java
import java.util.*;public class Main { public static int maximumConsecutiveNumbers(int[] arr) { // Store all non-zero elements // in a new List and // calculate total zero elements List<Integer> v = new ArrayList<>(); // Variable to store the // count of zero elements int count = 0; for (int i = 0; i < arr.length; i++) { if (arr[i] == 0) { count++; } else { v.add(arr[i]); } } // Sort the array Collections.sort(v); // Remove all the duplicates // from the array for (int i = 1; i < v.size(); i++) { if (v.get(i) == v.get(i - 1)) { v.remove(i); i--; } } // Variable to store the maximum // value of the sequence int MAXN = 1100000; // Make the prefix array int[] pref = new int[MAXN + 1]; Arrays.fill(pref, 0); for (int i = 0; i < v.size(); i++) { pref[v.get(i)]++; } for (int i = 1; i <= MAXN; i++) { pref[i] += pref[i - 1]; } int mx = 0; // Iterate for each element and // use for (int i = 1; i <= MAXN; i++) { int l = i, r = MAXN; int localMax = 0; while (l <= r) { int mid = (l + r) / 2; // Conversions equal to number // of zeros, can be made upto mid if (pref[mid] - pref[i - 1] + count >= (mid - i + 1)) { l = mid + 1; localMax = Math.max(localMax, mid - i + 1); } else { r = mid - 1; } } mx = Math.max(mx, localMax); } return mx; } public static void main(String[] args) { int N = 7; int arr[] = { 0, 6, 5, 10, 3, 0, 11 }; System.out.println(maximumConsecutiveNumbers(arr)); }} |
Python3
# Python 3 implementation for the above approach# Function to calculate maximum# possible consecutive numbers# with changes alloweddef maximumConsecutiveNumbers(arr, N): # Store all non-zero elements # in a new vector and # calculate total zero elements v = [] # Variable to store the # count of zero elements count = 0 for i in range(N): if(arr[i] == 0): count += 1 else: v.append(arr[i]) # Sort the array v.sort() # Remove all the duplicates # from the array v = set(v) v = list(v) # Variable to store the maximum # value of the sequence MAXN = 110000 # Make the prefix array pref = [0 for i in range(MAXN+1)] for i in range(len(v)): pref[v[i]] += 1 for i in range(1,MAXN+1,1): pref[i] += pref[i - 1] mx = 0 # Iterate for each element and # use binary search for i in range(1,MAXN+1,1): l = i r = MAXN local_max = 0 while (l <= r): mid = (l + r) // 2 # Conversions equal to number # of zeros, can be made upto mid if (pref[mid] - pref[i - 1] + count >= (mid - i + 1)): l = mid + 1 local_max = max(local_max,mid - i + 1) else: r = mid - 1 mx = max(mx, local_max) return mx# Driver Codeif __name__ == '__main__': N = 7 arr = [0, 6, 5, 10, 3, 0, 11] print(maximumConsecutiveNumbers(arr, N)) # This code is contributed by ipg2016107. |
C#
// C# implementation for the above approachusing System;using System.Collections.Generic;using System.Linq;class GFG { // Function to calculate maximum // possible consecutive numbers // with changes allowed static int maximumConsecutiveNumbers(int[] arr, int N) { // Store all non-zero elements // in a new vector and // calculate total zero elements List<int> v = new List<int>(); // Variable to store the // count of zero elements int count = 0; for (int i = 0; i < N; i++) { if (arr[i] == 0) { count++; } else { v.Add(arr[i]); } } // Sort the array v.Sort(); // Remove all the duplicates // from the array List<int> distinct = v.Distinct().ToList(); // Variable to store the maximum // value of the sequence int MAXN = 1100000; // Make the prefix array List<int> pref = new List<int>(new int[MAXN + 1]); for (int i = 0; i < distinct.Count; i++) { pref[distinct[i]]++; } for (int i = 1; i <= MAXN; i++) { pref[i] += pref[i - 1]; } int mx = 0; // Iterate for each element and // use binary search for (int i = 1; i <= MAXN; i++) { int l = i, r = MAXN; int local_max = 0; while (l <= r) { int mid = (l + r) / 2; // Conversions equal to number // of zeros, can be made upto mid if (pref[mid] - pref[i - 1] + count >= (mid - i + 1)) { l = mid + 1; local_max = Math.Max(local_max, mid - i + 1); } else { r = mid - 1; } } mx = Math.Max(mx, local_max); } return mx; } // Driver Code public static void Main() { int N = 7; int[] arr = { 0, 6, 5, 10, 3, 0, 11 }; Console.Write(maximumConsecutiveNumbers(arr, N)); }}// This code is contributed by ukasp. |
Javascript
<script> // Javascript implementation for the above approach // Function to calculate maximum // possible consecutive numbers // with changes allowed const maximumConsecutiveNumbers = (arr, N) => { // Store all non-zero elements // in a new vector and // calculate total zero elements let v = []; // Variable to store the // count of zero elements let count = 0; for (let i = 0; i < N; i++) { if (arr[i] == 0) { count++; } else { v.push(arr[i]); } } // Sort the array // Remove all the duplicates // from the array v = [...new Set(v)]; // Variable to store the maximum // value of the sequence let MAXN = 1100000; // Make the prefix array let pref = new Array(MAXN + 1).fill(0); for (let i = 0; i < v.length; i++) { pref[v[i]]++; } for (let i = 1; i <= MAXN; i++) { pref[i] += pref[i - 1]; } let mx = 0; // Iterate for each element and // use binary search for (let i = 1; i <= MAXN; i++) { let l = i, r = MAXN; let local_max = 0; while (l <= r) { let mid = parseInt((l + r) / 2); // Conversions equal to number // of zeros, can be made upto mid if (pref[mid] - pref[i - 1] + count >= (mid - i + 1)) { l = mid + 1; local_max = Math.max(local_max, mid - i + 1); } else { r = mid - 1; } } mx = Math.max(mx, local_max); } return mx; } // Driver Code let N = 7; let arr = [0, 6, 5, 10, 3, 0, 11]; document.write(maximumConsecutiveNumbers(arr, N)) // This code is contributed by rakeshsahni</script> |
Output
5
Time Complexity: O(N*log(K)), where K is maximum value in the array
Auxiliary Space: O(N)
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