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++
// 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 indicesvoid 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 Codeint main(){ int arr[] = { 0, 1, 2, 3, 5 }; int N = sizeof(arr) / sizeof(arr[0]); smlstNonNeg(arr, N);} |
Java
// Java program to implement// the above approachimport java.io.*;import java.util.Arrays;class GFG{ // Function to print the smallest// missing non-negative integer// up to every array indicesstatic 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 Codepublic 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 |
Python3
# Python3 program to implement# the above approach# Function to print smallest# missing non-negative integer# up to every array indicesdef 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 Codeif __name__ == '__main__': arr = [ 0, 1, 2, 3, 5 ] N = len(arr) smlstNonNeg(arr, N)# This code is contributed by mohit kumar 29 |
C#
// C# program to implement// the above approachusing System; class GFG{ // Function to print the smallest// missing non-negative integer// up to every array indicesstatic 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 Codepublic static void Main (){ int[] arr = { 0, 1, 2, 3, 5 }; int N = arr.Length; smlstNonNeg(arr, N);}}// This code is contributed by code_hunt |
Javascript
<script>// Javascript program to implement// the above approach// Function to print the smallest// missing non-negative integer// up to every array indicesfunction smlstNonNeg(arr, N){ // Stores the smallest missing // non-negative integers between // start index to current index let smNonNeg = 0; // Store the boolean value to check // smNonNeg present between start // index to each index of the array let hash = []; for(let i = 0; i < N; i++) { hash[i] = false; } // Traverse the array for(let 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 document.write(smNonNeg + " "); }}// Driver Codelet arr = [ 0, 1, 2, 3, 5 ];let N = arr.length; smlstNonNeg(arr, N);// This code is contributed by target_2 </script> |
Output:
1 2 3 4 4
Time Complexity: O(N)
Auxiliary Space: O(N)
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