1D Array Problems:

1. Extreme Printing - Two Pointer

Approach: Use two pointers from start and end

void ExtremePrint(int arr[], int size) {
    for(int i=0;i<size/2;i++) {
        cout <<arr[i]<<" "<<arr[size - 1 - i]<<" ";
    }
    if (size%2!= 0) {
        cout<<arr[size / 2]; // Print the middle element if the array size is odd
    }
    cout << endl;
}

2. Count 0 and 1

Method : Count number of zeros and ones, then fill the array

void countZerosOnes(vector<int>& arr) {
    int oneCount = 0;
    int zeroCount = 0;
    int twoCount = 0; // Just for the logic there is no need to create twoCount Variable

    for (int i = 0; i < n; i++) {
        if (arr[i] == 0) {
            zeroCount++;
        } 
        else if (arr[i] == 1) {
            oneCount++;
        }
        else {
            twoCount++; // Just for the logic there is no need to create twoCount Variable
        }
    }

    fill(arr, arr + zeroCount, 0);
    fill(arr + zeroCount, arr + zeroCount + oneCount, 1);
    fill(arr + zeroCount + oneCount, arr + n, 2);

}

3. Min Max in Array

Approach: Use INT_MIN and INT_MAX, traverse the array

void minandmax(int arr[],int size){
  int min =INT_MAX;
  int max= INT_MIN;

  for(int i=0;i<size;i++){
    if(arr[i]<min) min=arr[i];

    if(arr[i]>max) max=arr[i];
  }

  cout<<"Min elem :"<<min<<" Max Elem :"<<max ;
  cout<<endl;

}

4. Linear Search

Approach: Traverse array to find target element

int findElement(int arr[], int size, int elemt) {
    for(int i = 0; i < size; i++) {
        if(elemt == arr[i]) {
            cout << "Element found" << endl;
            return i; // Return index if element is found
        }
    }
    cout << "Element not in the array" << endl;
    return -1; // Return -1 if element is not found
}

5. LeetCode 136: Single Number

Approach: XOR approach - all duplicates cancel out

class Solution {
public:
    int singleNumber(vector<int>& nums) {
        int ans = 0;
        
        for(int i = 0; i < nums.size(); i++){
            ans = ans ^ nums[i];
        }
        
        return ans;
    }
};

6. Sort 0 and 1

Method 1: Count and fill

Method 2: Sort function

Method 3: Two pointer approach