[Array] Find Target Indices After Sorting Array - Easy

Finding target indices after sorting an array means we need to find where a certain target value is in a sorted array. First, we sort the array. Then we look through it to find the positions where the target value shows up. This problem is important in making algorithms and we can do it well in many programming languages.

In this article, we will look at the basics of finding target indices after sorting an array. We will begin with the problem statement. Then, we will provide detailed examples in Java, Python, and C++. We will also talk about ways to make the process faster, special cases we might face, and how to analyze the complexity. We will give practical examples to show the solution. Lastly, we will add a section with frequently asked questions to help clear up common doubts about target indices in sorted arrays.

  • Array Target Indices After Sorting Array Solution Overview
  • Understanding the Problem Statement for Finding Target Indices
  • Java Approach to Finding Target Indices After Sorting Array
  • Python Implementation of Finding Target Indices After Sorting Array
  • C++ Solution for Finding Target Indices After Sorting Array
  • Optimizing the Search for Target Indices in Sorted Array
  • Handling Edge Cases in Finding Target Indices
  • Complexity Analysis of Finding Target Indices
  • Practical Examples of Finding Target Indices After Sorting
  • Frequently Asked Questions

For more reading on similar array problems, we can check out articles like Array Two Sum and Array Best Time to Buy and Sell Stock.

Understanding the Problem Statement for Finding Target Indices

To find target indices after sorting an array, we need to understand the problem statement well. Our task is to find the indices of a certain target value in a sorted version of a given array.

Problem Breakdown:

  • Input: An integer array nums and an integer target.
  • Output: A list of indices where the target is in the sorted version of nums.

Example:

If we have this input:

nums = [4, 2, 7, 1, 3]
target = 2

After we sort nums, we get:

sorted_nums = [1, 2, 3, 4, 7]

The target 2 is at index 1 in sorted_nums. So, the output is:

output = [1]

Constraints:

  • The array can have duplicate values. We must consider all occurrences of the target.
  • If we do not find the target, we return an empty list.

This problem needs us to sort and search in an efficient way. We will look at how to do this in different programming languages in the next sections.

Java Approach to Finding Target Indices After Sorting Array

We can find target indices after sorting an array in Java. First, we need to sort the array. Then we will find the indices of the target element. Here is a simple way to do this using Java.

  1. Sort the Array: We use Arrays.sort() to sort the array.
  2. Find Target Indices: We loop through the sorted array to get the indices of the target value.

Java Code Implementation

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class TargetIndices {
    public static List<Integer> findTargetIndices(int[] nums, int target) {
        Arrays.sort(nums);
        List<Integer> indices = new ArrayList<>();
        
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == target) {
                indices.add(i);
            }
        }
        
        return indices;
    }

    public static void main(String[] args) {
        int[] nums = {1, 2, 5, 2, 3};
        int target = 2;
        List<Integer> targetIndices = findTargetIndices(nums, target);
        System.out.println("Target Indices: " + targetIndices);
    }
}

Explanation of the Code

  • Sorting: We sort the array nums in increasing order. This helps us to find the target indices easily.
  • Index Collection: We go through the sorted array. We check for the target value. When we find it, we add the index to the list indices.
  • Output: In the end, we print the target indices. This shows their places in the sorted array.

Example

If we have the input array [1, 2, 5, 2, 3] and the target is 2, the output will be:

Target Indices: [1, 2]

This means that the target 2 is at indices 1 and 2 after sorting the array.

This Java way sorts the array and finds the target indices in a clear way. For more problems with arrays, look at Array Two Sum or Array Rotate Array.

Python Implementation of Finding Target Indices After Sorting Array

We can find the target indices of a value in a sorted array with a simple method in Python. First, we sort the array. Then we look for the indices of the target value.

Here is a simple code in Python:

def target_indices(arr, target):
    # Sort the array
    arr.sort()
    # Find target indices
    indices = [i for i, x in enumerate(arr) if x == target]
    return indices

# Example usage
arr = [1, 2, 5, 2, 3]
target = 2
print(target_indices(arr, target))  # Output: [1, 2]

Explanation of the Code:

  1. Sorting: The sort() method will sort the array in increasing order.
  2. Finding Indices: We use a list comprehension with enumerate() to get indices where the element is the same as the target.

Complexity Analysis:

  • Sorting the array takes (O(n n)). Here (n) is the number of elements in the array.
  • Finding indices takes (O(n)).
  • Overall, the complexity is (O(n n)).

This method helps us to find all target indices in the sorted array. It gives correct and fast results. If we want to learn more about similar problems with arrays, we can check Array Two Sum or Array Contains Duplicate for more help.

C++ Solution for Finding Target Indices After Sorting Array

We can find target indices after sorting an array in C++ by using a simple way. First, we sort the array. Then, we look through the sorted array to find where the target element is. Below is a simple example.

#include <iostream>
#include <vector>
#include <algorithm>

std::vector<int> targetIndices(std::vector<int>& nums, int target) {
    std::vector<int> indices;
    std::sort(nums.begin(), nums.end());

    for (int i = 0; i < nums.size(); i++) {
        if (nums[i] == target) {
            indices.push_back(i);
        }
    }

    return indices;
}

int main() {
    std::vector<int> nums = {1, 2, 5, 2, 3};
    int target = 2;
    std::vector<int> result = targetIndices(nums, target);

    std::cout << "Target Indices: ";
    for (int index : result) {
        std::cout << index << " ";
    }

    return 0;
}

Explanation of the Code:

We include some important headers like <iostream>, <vector>, and <algorithm>. The function targetIndices sorts the input vector nums. Then, it goes through the vector to find the indices of the target.

The indices where target appears are saved in the indices vector. We return this vector at the end. The main function shows how we can use the targetIndices function and print the indices we found.

Performance:

The sorting step takes time O(n log n). Scanning the array takes time O(n). So the total time is O(n log n). This way, we can find all indices of the target in the sorted array.

If you want to learn more about similar problems and solutions, check these articles: Array Two Sum and Array Contains Duplicate.

Optimizing the Search for Target Indices in Sorted Array

We can make searching for target indices in a sorted array better by using binary search. Since the array is sorted, we can find the target quickly with a simple binary search method.

Binary Search Approach

  1. Initialization: We start with two pointers. One called left at the beginning of the array and another called right at the end.

  2. Binary Search Logic:

    • We calculate the middle index: mid = left + (right - left) / 2.
    • We compare the middle element with the target:
      • If array[mid] is equal to the target, we save the index. Then we keep searching in the left side for any duplicates.
      • If array[mid] is less than the target, we move the left pointer to mid + 1.
      • If array[mid] is greater than the target, we move the right pointer to mid - 1.

  3. Collecting Indices: After we find the target, we go through the sorted array to collect all the indices where the target appears.

Implementation Example

Java Implementation

import java.util.ArrayList;
import java.util.List;

public class TargetIndices {
    public List<Integer> targetIndices(int[] nums, int target) {
        List<Integer> result = new ArrayList<>();
        Arrays.sort(nums);
        int left = 0, right = nums.length - 1;

        while (left <= right) {
            int mid = left + (right - left) / 2;

            if (nums[mid] == target) {
                result.add(mid);
                // Check for duplicates on the left
                for (int i = mid - 1; i >= left && nums[i] == target; i--) {
                    result.add(i);
                }
                // Check for duplicates on the right
                for (int i = mid + 1; i <= right && nums[i] == target; i++) {
                    result.add(i);
                }
                break;
            } else if (nums[mid] < target) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        }
        return result;
    }
}

Python Implementation

def target_indices(nums, target):
    nums.sort()
    left, right = 0, len(nums) - 1
    result = []
    
    while left <= right:
        mid = left + (right - left) // 2
        
        if nums[mid] == target:
            result.append(mid)
            # Check for duplicates on the left
            i = mid - 1
            while i >= left and nums[i] == target:
                result.append(i)
                i -= 1
            # Check for duplicates on the right
            i = mid + 1
            while i <= right and nums[i] == target:
                result.append(i)
                i += 1
            break
        elif nums[mid] < target:
            left = mid + 1
        else:
            right = mid - 1
            
    return sorted(result)

Complexity Analysis

  • Time Complexity: O(n log n) because we sort the array first. After that, we have O(log n) for binary search and O(k) for collecting indices. Here k is the number of times the target appears.
  • Space Complexity: O(k) for the list of indices we return.

This better approach reduces the number of comparisons. It uses the sorted nature of the array, which helps a lot when we deal with bigger datasets.

For more related problems, you can check these articles: Array Two Sum and Array Maximum Subarray.

Handling Edge Cases in Finding Target Indices

When we make a solution for finding target indices after sorting an array, we need to think about some edge cases. These cases can change how accurate and fast our algorithm is. Here are the main edge cases to handle:

  1. Empty Array:
    • If the input array is empty, we should return an empty list. There are no elements to sort or check.
    if (nums.length == 0) {
    return new ArrayList<>();
}
  2. Single Element Array:
    • For an array with only one element, if the target is the same as that element, we return the index 0. If not, we return an empty list.
    if (nums.length == 1) {
    return nums[0] == target ? List.of(0) : List.of();
}
  3. Target Not Present:
    • If the target is not in the array, we must return an empty list. We check this after sorting and searching.
  4. Multiple Occurrences of Target:
    • If the target is in the array many times, we should return all indices where the target is found. We must collect all indices while we search.
  5. Negative or Zero Values:
    • Arrays can have negative values or zeros. Our algorithm should work with these without any special changes. Sorting will arrange them right.
  6. All Elements Same:
    • If all elements in the array are the same and equal to the target, we return all indices. The algorithm must handle this well without going through extra steps.
  7. Large Arrays:
    • For very big arrays, we need to make sure our algorithm can handle this size on time. We might use better data structures or algorithms to keep the time low.
  8. Out of Bound Target:
    • If the target value is smaller than the smallest or bigger than the biggest value in the sorted array, we should return an empty list.

We should include these edge cases in our implementation to make sure it works well. By thinking about these scenarios, we can make a good solution for finding target indices after sorting an array.

Example in Python

Here is a Python example that thinks about these edge cases:

def target_indices(nums, target):
    nums.sort()
    indices = []
    for i, num in enumerate(nums):
        if num == target:
            indices.append(i)
    return indices

This example sorts the array and then collects indices of the target value. It manages the edge cases we talked about above.

Complexity Analysis of Finding Target Indices

When we look at the complexity of finding target indices after we sort an array, we have two main steps. First, we sort the array. Then, we search for the target indices.

  1. Sorting Complexity:
    • Sorting usually takes O(n log n) time. Here, n is the number of elements in the array. This happens because sorting methods like QuickSort or MergeSort are efficient.
  2. Searching Complexity:
    • After we sort, we can find the target indices in linear time O(n). We do this by going through the sorted array and checking for the target. If we use a binary search, we can make the time faster to O(log n).
  3. Overall Complexity:
    • The total time complexity to find target indices after sorting the array is O(n log n). This is because sorting takes the most time. The space complexity is O(1) if we sort in place. If we create a new array for the sorted output, the space complexity is O(n).

Example of Complexity

Let’s take an array arr = [3, 1, 2, 4, 1] and a target value target = 1: - When we sort the array, we get sorted_arr = [1, 1, 2, 3, 4]. This takes O(n log n). - To find the target indices for 1 in the sorted array, we can do it in O(n). We get the indices [0, 1].

This analysis shows how important the sorting step is for the overall speed of finding target indices in an array.

Practical Examples of Finding Target Indices After Sorting

We will show how to find target indices after sorting an array. We will use some simple examples in Java, Python, and C++.

Example 1: Java

We will sort the array and find the target indices in this example.

import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;

public class TargetIndices {
    public static List<Integer> targetIndices(int[] nums, int target) {
        Arrays.sort(nums);
        List<Integer> result = new ArrayList<>();
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == target) {
                result.add(i);
            }
        }
        return result;
    }

    public static void main(String[] args) {
        int[] nums = {3, 2, 3, 1, 2, 4};
        int target = 2;
        System.out.println(targetIndices(nums, target));  // Output: [1, 2]
    }
}

Example 2: Python

Now we will do the same thing in Python.

def target_indices(nums, target):
    nums.sort()
    return [i for i, num in enumerate(nums) if num == target]

if __name__ == "__main__":
    nums = [3, 2, 3, 1, 2, 4]
    target = 2
    print(target_indices(nums, target))  # Output: [1, 2]

Example 3: C++

In C++, we can do the same work with this code.

#include <iostream>
#include <vector>
#include <algorithm>

std::vector<int> targetIndices(std::vector<int>& nums, int target) {
    std::sort(nums.begin(), nums.end());
    std::vector<int> result;
    for (int i = 0; i < nums.size(); i++) {
        if (nums[i] == target) {
            result.push_back(i);
        }
    }
    return result;
}

int main() {
    std::vector<int> nums = {3, 2, 3, 1, 2, 4};
    int target = 2;
    std::vector<int> indices = targetIndices(nums, target);
    for (int index : indices) {
        std::cout << index << " ";  // Output: 1 2 
    }
    return 0;
}

Example Explanation

  • Sorting: First, we sort the array in each example. This helps us to get the correct order.
  • Finding Indices: After we sort, we check the sorted array to find the indices of the target value.
  • Output: The output shows where the target value is after sorting the array.

These simple examples show how we can find target indices after sorting an array in Java, Python, and C++. For more problems related to arrays, we can look at sites like Array Two Sum or Array Maximum Subarray.

Frequently Asked Questions

What is the problem statement for finding target indices after sorting an array?

The problem of finding target indices after sorting an array is about figuring out where a specific target value is located in the array after we sort it. This is important for many uses like searching and indexing. When we sort the array and see where the target value is, we can easily get its indices. If you want to learn more, check our article on Array Target Indices After Sorting Array.

How can I implement the target indices search in Java?

To do target indices search in Java, we first sort the array using Arrays.sort(). Then, we can use a loop to look at each element and check if it is the target value. We save the indices of where we find the target in a list. This way, we can find the correct positions after sorting. For examples and code, look at our section on Java Approach to Finding Target Indices.

What Python techniques are best for finding target indices after sorting?

In Python, we can use the built-in sorted() function to sort the array. After that, we can use a list comprehension or a loop to find and collect the indices of the target value. This method is simple and takes advantage of Python’s good built-in tools. For a full implementation, see our Python Implementation of Finding Target Indices.

Are there any edge cases to consider when finding target indices?

Yes, when we find target indices after sorting an array, we should think about edge cases like an empty array, an array without the target, or if the target appears many times. These situations can change the output and we need to handle them well in our code. For more on this, look at our section on Handling Edge Cases in Finding Target Indices.

To analyze time complexity for finding target indices, we need to think about the sorting step and the searching step. Sorting an array usually takes (O(n n)) time, and searching for indices can take (O(n)). So, the total time is mainly affected by the sorting step, resulting in (O(n n)). For a detailed breakdown, visit our Complexity Analysis of Finding Target Indices section.