Smallest missing non-negative integer upto every array index

Given an array arr[] of size N, the task is for every array indices is to find the smallest missing non-negative integer upto that index of the given array.

Examples:

Input: arr[] = {1, 3, 0, 2}
Output: 0 0 2 4
Explanation:
Smallest missing non-negative integer from index 0 to 0 is 0.
Smallest missing non-negative integer from index 0 to 1 is 0.
Smallest missing non-negative integer from index 0 to 2 is 2.
Smallest missing non-negative integer from index 0 to 3 is 4.

Input: arr[] = {0, 1, 2, 3, 5}
Output: 1 2 3 4 4

 

Approach: This problem can be solved using Hashing. Follow the steps below to solve the problem:



  • Initialize a variable, say smNonNeg to store the smallest missing non-negative integers between the start index and the current index of the given array.
  • Initialize an array, say hash[N] to check if smNonNeg present between the start index and the current index or not.
  • Traverse the given array and check if hash[smNonNeg] equal to 0 or not. If found to be true, then print the value of smNonNeg.
  • Otherwise, increment the value of smNonNeg while hash[smNonNeg] not equal to 0.

Below is the implementation of the above approach:

C++

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// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to print the smallest
// missing non-negative integer
// up to every array indices
void smlstNonNeg(int arr[], int N)
{
    // Stores the smallest missing
    // non-negative integers between
    // start index to current index
    int smNonNeg = 0;
 
    // Store the boolean value to check
    // smNonNeg present between start
    // index to each index of the array
    bool hash[N + 1] = { 0 };
 
    // Traverse the array
    for (int i = 0; i < N; i++) {
 
        // Since output always lies
        // in the range[0, N - 1]
        if (arr[i] >= 0 and arr[i] < N) {
            hash[arr[i]] = true;
        }
 
        // Check if smNonNeg is
        // present between start index
        // and current index or not
        while (hash[smNonNeg]) {
            smNonNeg++;
        }
 
        // Print smallest missing
        // non-negative integer
        cout << smNonNeg << " ";
    }
}
 
// Driver Code
int main()
{
    int arr[] = { 0, 1, 2, 3, 5 };
    int N = sizeof(arr) / sizeof(arr[0]);
    smlstNonNeg(arr, N);
}

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Java

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// Java program to implement
// the above approach
import java.io.*;
import java.util.Arrays;
 
class GFG{
  
// Function to print the smallest
// missing non-negative integer
// up to every array indices
static void smlstNonNeg(int arr[], int N)
{
     
    // Stores the smallest missing
    // non-negative integers between
    // start index to current index
    int smNonNeg = 0;
  
    // Store the boolean value to check
    // smNonNeg present between start
    // index to each index of the array
    Boolean[] hash = new Boolean[N + 1];
    Arrays.fill(hash, false);
 
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // Since output always lies
        // in the range[0, N - 1]
        if (arr[i] >= 0 && arr[i] < N)
        {
            hash[arr[i]] = true;
        }
  
        // Check if smNonNeg is
        // present between start index
        // and current index or not
        while (hash[smNonNeg])
        {
            smNonNeg++;
        }
  
        // Print smallest missing
        // non-negative integer
        System.out.print(smNonNeg + " ");
    }
}
  
// Driver Code
public static void main (String[] args)
{
    int arr[] = { 0, 1, 2, 3, 5 };
    int N = arr.length;
     
    smlstNonNeg(arr, N);
}
}
 
// This code is contributed by sanjoy_62

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Python3

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# Python3 program to implement
# the above approach
 
# Function to prthe smallest
# missing non-negative integer
# up to every array indices
def smlstNonNeg(arr, N):
     
    # Stores the smallest missing
    # non-negative integers between
    # start index to current index
    smNonNeg = 0
 
    # Store the boolean value to check
    # smNonNeg present between start
    # index to each index of the array
    hash = [0] * (N + 1)
 
    # Traverse the array
    for i in range(N):
 
        # Since output always lies
        # in the range[0, N - 1]
        if (arr[i] >= 0 and arr[i] < N):
            hash[arr[i]] = True
 
        # Check if smNonNeg is
        # present between start index
        # and current index or not
        while (hash[smNonNeg]):
            smNonNeg += 1
 
        # Print smallest missing
        # non-negative integer
        print(smNonNeg, end = " ")
 
# Driver Code
if __name__ == '__main__':
     
    arr = [ 0, 1, 2, 3, 5 ]
    N = len(arr)
     
    smlstNonNeg(arr, N)
 
# This code is contributed by mohit kumar 29

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C#

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// C# program to implement
// the above approach
using System;
  
class GFG{
   
// Function to print the smallest
// missing non-negative integer
// up to every array indices
static void smlstNonNeg(int[] arr, int N)
{
     
    // Stores the smallest missing
    // non-negative integers between
    // start index to current index
    int smNonNeg = 0;
   
    // Store the boolean value to check
    // smNonNeg present between start
    // index to each index of the array
    bool[] hash = new bool[N + 1];
     
    for(int i = 0; i < N; i++)
    {
        hash[i] = false;
    }
     
    // Traverse the array
    for(int i = 0; i < N; i++)
    {
         
        // Since output always lies
        // in the range[0, N - 1]
        if (arr[i] >= 0 && arr[i] < N)
        {
            hash[arr[i]] = true;
        }
   
        // Check if smNonNeg is
        // present between start index
        // and current index or not
        while (hash[smNonNeg])
        {
            smNonNeg++;
        }
   
        // Print smallest missing
        // non-negative integer
        Console.Write(smNonNeg + " ");
    }
}
   
// Driver Code
public static void Main ()
{
    int[] arr = { 0, 1, 2, 3, 5 };
    int N = arr.Length;
      
    smlstNonNeg(arr, N);
}
}
 
// This code is contributed by code_hunt

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Output: 

1 2 3 4 4









 

Time Complexity: O(N)
Auxiliary Space: O(N)

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