[Array] Replace Elements with Greatest Element on Right Side - Easy

In the problem of replacing elements with the biggest element on the right side of an array, we need to go through the array. For each element, we replace it with the highest value we find to its right. The last element of the array gets replaced with -1 because there are no elements on its right. We can solve this task in a good way by moving from right to left in the array. This way, we check each element just once. This gives us a time complexity of O(n).

In this article, we will look closely at the idea of replacing elements with the greatest element on the right side. We will begin by understanding the problem. Then, we will show the best solutions using right-to-left techniques in Java, Python, and C++. We will share code examples for each implementation. Finally, we will compare the different methods and answer common questions.

  • [Array] Replace Elements with Greatest Element on Right Side - A Simple Guide
  • Understanding the Problem for Array Replacement
  • Best Solution Using Right to Left Traversal in Java
  • Java Code for Replace Elements with Greatest Element
  • Best Solution Using Right to Left Traversal in Python
  • Python Code for Replace Elements with Greatest Element
  • Best Solution Using Right to Left Traversal in C++
  • C++ Code for Replace Elements with Greatest Element
  • Comparing Different Methods
  • Common Questions

For more information on array problems, we can check these articles: Array Two Sum - Easy, Array Best Time to Buy and Sell Stock - Easy, and Array Contains Duplicate - Easy.

[Array] Replace Elements with Greatest Element on Right Side - A Simple Guide

Understanding the Problem for Array Replacement

The problem for “Replace Elements with Greatest Element on Right Side” is simple. We have an array of numbers. We need to change each number with the biggest number to its right. The last number will be -1 because there are no numbers on its right.

Problem Requirements:

  • Input: An array of numbers arr.
  • Output: A new array where each number at index i is changed by the biggest number in arr[i+1] to arr[n-1].
  • Edge Case: If there are no numbers to the right of the last number, we change it to -1.

Example:

  • Input: [17, 18, 5, 4, 6, 1]
  • Output: [18, 6, 6, 6, 1, -1]

Approach:

  1. We will go through the array from right to left.
  2. We will keep track of the biggest number we found so far.
  3. We will change the current number with this biggest number.
  4. We will update the biggest number as needed.

This way, we only go through the array one time. This gives us a good O(n) time and O(1) space if we change the array directly.

If we want to learn more about similar array problems, we can check articles like Array Two Sum or Array Best Time to Buy and Sell Stock.

Optimal Solution Using Right to Left Traversal in Java

We can replace each element in an array with the greatest element on its right side using a right-to-left traversal. This method helps us check each element just once. So, we get a time complexity of O(n) and a space complexity of O(1).

Algorithm Steps:

  1. We start by setting a variable max to -1. This will hold the greatest element we find on the right of the current element.
  2. Then, we go through the array from the last element to the first one.
  3. For each element, we save its value, replace it with max, and update max to be the greater of the current element and the previous value of max.

Java Implementation:

public class ReplaceElements {
    public static int[] replaceElements(int[] arr) {
        int n = arr.length;
        int max = -1; // We set max to -1 for the last element
        for (int i = n - 1; i >= 0; i--) {
            int current = arr[i]; // Save current element
            arr[i] = max; // Replace it with max
            max = Math.max(max, current); // Update max
        }
        return arr;
    }

    public static void main(String[] args) {
        int[] arr = {17, 18, 5, 4, 6, 1};
        int[] result = replaceElements(arr);
        for (int num : result) {
            System.out.print(num + " "); // Output: 18 6 6 6 1 -1
        }
    }
}

Explanation:

  • We start the loop from the end of the array and go backward. This way, each element gets replaced with the greatest element we found to its right.
  • The last element always gets replaced with -1 because there are no elements to its right.
  • This method is simple and works well, making it a good choice for problems with maximum elements in arrays.

For more about how to work with arrays, we can read about Array Two Sum or Array Maximum Subarray.

Java Implementation of Replace Elements with Greatest Element

To solve the problem of replacing each element in an array with the greatest element on its right side in Java, we can use a right-to-left approach. This way, we can easily find the maximum value on the right side as we go through the array.

Code Implementation

Here is a simple Java code for the algorithm:

public class ReplaceElements {
    public int[] replaceElements(int[] arr) {
        int n = arr.length;
        int maxFromRight = -1;  // We start maxFromRight at -1

        for (int i = n - 1; i >= 0; i--) {
            int current = arr[i]; // Save the current element
            arr[i] = maxFromRight; // Change current element to maxFromRight
            maxFromRight = Math.max(maxFromRight, current); // Update maxFromRight
        }

        return arr;
    }

    public static void main(String[] args) {
        ReplaceElements re = new ReplaceElements();
        int[] arr = {17, 18, 5, 4, 6, 1};
        int[] result = re.replaceElements(arr);
        System.out.println(Arrays.toString(result)); // Output: [18, 6, 6, 6, 1, -1]
    }
}

Explanation of the Code

  • Initialization: We begin with maxFromRight set to -1 because the rightmost element has no greater element.
  • Traversal: We go from the last index to the first index of the array.
  • Replacement: In each step, we replace the current element with maxFromRight. Then, we update maxFromRight to be the bigger of itself and the current element.
  • Output: We return the modified array, where each element is now replaced with the greatest element to its right.

Time Complexity

  • The time complexity of this algorithm is O(n) because we go through the array only one time.

Space Complexity

  • The space complexity is O(1) since we change the array in place without using extra data structures.

This method helps us meet the problem’s needs while keeping good performance. For more similar array problems, you can check out Array: Maximum Subarray.

Optimal Solution Using Right to Left Traversal in Python

To solve the problem of replacing elements in an array with the greatest element on their right side, we can use a right-to-left traversal. This way, we only go through the array one time. This gives us a time complexity of O(n) and a space complexity of O(1).

Steps:

  1. Initialization: We make a variable called max_from_right to keep track of the highest value found from the right side of the array.
  2. Traversal: We start at the last element and go to the first element. For every element, we save the current value, replace the current element with max_from_right, and then update max_from_right with the bigger value between the current value and max_from_right.
  3. Edge Case: The last element should be replaced with -1 since there are no elements on its right.

Python Code Implementation:

def replaceElements(arr):
    max_from_right = -1
    n = len(arr)
    
    for i in range(n - 1, -1, -1):
        current = arr[i]
        arr[i] = max_from_right
        max_from_right = max(max_from_right, current)
    
    return arr

# Example usage:
array = [17, 18, 5, 4, 6, 1]
result = replaceElements(array)
print(result)  # Output: [18, 6, 6, 6, 1, -1]

This code replaces each element with the highest element to its right. It follows the problem rules. Using just one loop makes this solution work well, even for big datasets.

If we want to learn more about working with arrays in Python, we can check out related topics like array maximum subarray or array contains duplicate.

Python Implementation of Replace Elements with Greatest Element

We can solve the problem of replacing elements in an array with the greatest element on their right side using Python. We will use a right-to-left method. This way, we can replace each element quickly, keeping the time to O(n).

The main idea is to go through the array from the end to the start. We will keep track of the highest value we have seen so far. Here is how we do it:

  1. Start from the last element. We will set a variable to track the maximum value. We can start it at -1 because we will use it to replace the last element.
  2. Go from the second last element to the first element in the array.
  3. For each element, we save its original value. We replace it with the current maximum value. Then we update the maximum value if the original value is bigger than the current maximum.

Here is the Python code for this:

def replaceElements(arr):
    max_value = -1
    for i in range(len(arr) - 1, -1, -1):
        original_value = arr[i]
        arr[i] = max_value
        max_value = max(max_value, original_value)
    return arr

# Example usage
input_array = [17, 18, 5, 4, 6, 1]
result = replaceElements(input_array)
print(result)  # Output: [18, 6, 6, 6, 1, -1]

This code replaces each element in the input array with the greatest element on its right. The last element becomes -1 since there are no elements to its right.

The algorithm works well and uses little extra space. It only needs a small amount of extra space. This makes it good for big arrays.

If we want to learn more about array problems, we can check out this article on the maximum subarray.

Optimal Solution Using Right to Left Traversal in C++

We can solve the problem of replacing elements in an array with the greatest element on their right side. We will use a right-to-left traversal method in C++. This way, we only need to go through the array one time. This gives us a time complexity of O(n) and space complexity of O(1). We can change the array directly without needing extra space.

Algorithm Steps:

  1. First, we set a variable max_from_right to -1. This will keep the biggest value we find on the right of the current element.
  2. Next, we go through the array from the last element to the first element:
    • We save the current element in a temporary variable.
    • We change the current element to max_from_right.
    • We update max_from_right to be the bigger value between the current element and max_from_right.
  3. When we finish going through the array, the first element will be -1. This is because there are no elements to its right.

C++ Implementation:

#include <iostream>
#include <vector>
using namespace std;

vector<int> replaceElements(vector<int>& arr) {
    int n = arr.size();
    int max_from_right = -1;

    for (int i = n - 1; i >= 0; --i) {
        int temp = arr[i];
        arr[i] = max_from_right;
        max_from_right = max(max_from_right, temp);
    }

    return arr;
}

int main() {
    vector<int> arr = {17, 18, 5, 4, 6, 1};
    vector<int> result = replaceElements(arr);

    for (int num : result) {
        cout << num << " ";
    }
    return 0;
}

Example:

For the input array {17, 18, 5, 4, 6, 1}, the output will be {18, 6, 6, 6, 1, -1}. Each element is now replaced by the biggest element found to its right. The last element is replaced by -1.

This method is simple and works well. It also lets us change the original array without needing more memory. If you want to learn more about array techniques, you can look at topics like Array - Maximum Subarray or Array - Next Greater Element I.

C++ Implementation of Replace Elements with Greatest Element

To solve the problem of replacing each element in an array with the greatest element on its right side in C++, we use a right-to-left approach. This method lets us go through the array one time. This gives us a fast O(n) time.

Here’s how the algorithm works:

  1. We start with a variable to track the maximum element. We set it to -1 because we will replace elements.
  2. We go through the array from the last element to the first.
  3. For each element, we save the current maximum value to replace it. Then we update the maximum value if the current element is bigger.

C++ Code Implementation

#include <iostream>
#include <vector>
using namespace std;

vector<int> replaceElements(vector<int>& arr) {
    int maxElement = -1; // Start with -1 for the last element's replacement
    for (int i = arr.size() - 1; i >= 0; i--) {
        int current = arr[i]; // Save current element
        arr[i] = maxElement;  // Replace current element with maxElement
        if (current > maxElement) {
            maxElement = current; // Update maxElement if current is bigger
        }
    }
    return arr;
}

int main() {
    vector<int> arr = {17, 18, 5, 4, 6, 1};
    vector<int> result = replaceElements(arr);
    
    cout << "Replaced Array: ";
    for (int num : result) {
        cout << num << " ";
    }
    cout << endl;
    return 0;
}

Explanation of the Code

  • We create a function replaceElements that takes a list of integers as input.
  • We set maxElement to -1. This will help us replace the last element in the array.
  • We loop from the last index to the first index of the array. We update each element to the maxElement value.
  • If the current element is bigger than maxElement, we change maxElement to this current element.
  • In the end, we show the changed array.

This method helps us to replace each element with the greatest element to its right. It follows the problem’s needs. If you want to learn more about array techniques, you can check articles like Array Two Sum or Array Maximum Subarray.

[Array] Replace Elements with Greatest Element on Right Side - A Simple Guide

Understanding the Problem for Array Replacement

We need to replace each element in an array with the biggest element on its right. The last element will be -1 because there are no elements to its right. Given an array arr, we want to make a new array result. In result[i], we will put the maximum value from arr[i+1] to arr[n-1].

Best Solution Using Right to Left Traversal in Java

The best way to solve this problem is to look at the array from right to left. We will keep track of the biggest element we have seen so far.

public class ReplaceElements {
    public int[] replaceElements(int[] arr) {
        int n = arr.length;
        int maxSoFar = -1;
        for (int i = n - 1; i >= 0; i--) {
            int current = arr[i];
            arr[i] = maxSoFar;
            maxSoFar = Math.max(maxSoFar, current);
        }
        return arr;
    }
}

Java Code for Replace Elements with Greatest Element

To use this Java code, create an instance of the ReplaceElements class and call the replaceElements method with an integer array.

public class Main {
    public static void main(String[] args) {
        ReplaceElements re = new ReplaceElements();
        int[] result = re.replaceElements(new int[]{17, 18, 5, 4, 6, 1});
        System.out.println(Arrays.toString(result)); // Output: [18, 6, 6, 6, 1, -1]
    }
}

Best Solution Using Right to Left Traversal in Python

The Python code works in the same way. We keep track of the biggest value we see during the right-to-left loop.

def replaceElements(arr):
    max_so_far = -1
    for i in range(len(arr) - 1, -1, -1):
        current = arr[i]
        arr[i] = max_so_far
        max_so_far = max(max_so_far, current)
    return arr

Python Code for Replace Elements with Greatest Element

You can try this Python code by calling the replaceElements function with an example array.

result = replaceElements([17, 18, 5, 4, 6, 1])
print(result)  # Output: [18, 6, 6, 6, 1, -1]

Best Solution Using Right to Left Traversal in C++

The C++ code also uses the same right-to-left method, updating elements as we go through the array.

#include <vector>
#include <iostream>
using namespace std;

vector<int> replaceElements(vector<int>& arr) {
    int maxSoFar = -1;
    for (int i = arr.size() - 1; i >= 0; i--) {
        int current = arr[i];
        arr[i] = maxSoFar;
        maxSoFar = max(maxSoFar, current);
    }
    return arr;
}

C++ Code for Replace Elements with Greatest Element

To run the C++ code, create a main function to test the replaceElements function.

int main() {
    vector<int> arr = {17, 18, 5, 4, 6, 1};
    vector<int> result = replaceElements(arr);
    for (int val : result) {
        cout << val << " ";  // Output: 18 6 6 6 1 -1
    }
    return 0;
}

Comparing Different Approaches

The best solution using right-to-left loop is fast and clear. It has a time of O(n) and space of O(1):

  • Time Complexity: O(n) for going through the array once.
  • Space Complexity: O(1) because we change the input array directly.

Other ways, like using extra structures (like arrays or stacks) to hold maximums, would waste space. So, the right-to-left method is better for its speed and simplicity.

For more problems with arrays, you can check Array Two Sum and Array Maximum Subarray.

Frequently Asked Questions

1. What is the problem of replacing elements in an array with the greatest element on the right side?

The problem is about going through an array. We need to swap each element with the biggest value we find on its right. This task can help us learn more about handling arrays and making algorithms better. If you want to know more, look at Array: Maximum Subarray.

2. How can I solve the array replacement problem optimally?

To solve the array replacement problem in a good way, we can move from right to left. As we go through the array, we keep track of the biggest value we see. This way, we can change each element easily and we do not need extra space. This method works in O(n) time, which is good for big datasets.

3. What programming languages can I use to implement the solution?

We can use many programming languages for this solution. Some examples are Java, Python, and C++. Each language has different rules and ways to write code. But the main idea is the same. For example, you can look at our Java example for more details.

4. Are there any similar array problems I should be aware of?

Yes, there are many similar problems that can help us get better at coding. Some examples are the Two Sum Problem and the Best Time to Buy and Sell Stock. Trying these problems will help us learn different ways to handle arrays.

5. Can I apply this concept to other algorithmic challenges?

Sure! The idea of finding the biggest element on the right can be used in many other challenges. For example, we can find the next bigger element or work with subarrays. Knowing this will help us solve more problems. For example, look at the Next Greater Element I for a similar problem.