Merge pull request #4252 from AmruthaPariprolu/2460

2460

Author: ajay-dhangar

dsa-solutions/lc-solutions/2400-2499/2460-Apply-Operations-to-an-Array.md

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+---
+id:  Apply-Operations-to-an-Array
+title:  Apply Operations to an Array
+sidebar_label: 2460-Apply Operations to an Array
+tags: [dsa, leetcode]
+description: Problem solution of  Apply Operations to an Array
+---
+
+## Problem Statement 
+
+### Problem Description
+
+You are given a 0-indexed array nums of size n consisting of non-negative integers.
+
+You need to apply n - 1 operations to this array where, in the ith operation (0-indexed), you will apply the following on the ith element of nums:
+
+If nums[i] == nums[i + 1], then multiply nums[i] by 2 and set nums[i + 1] to 0. Otherwise, you skip this operation.
+After performing all the operations, shift all the 0's to the end of the array.
+
+For example, the array [1,0,2,0,0,1] after shifting all its 0's to the end, is [1,2,1,0,0,0].
+Return the resulting array.
+
+Note that the operations are applied sequentially, not all at once.
+ 
+### Examples
+
+#### Example 1
+```
+Input: nums = [1,2,2,1,1,0]
+Output: [1,4,2,0,0,0]
+Explanation: We do the following operations:
+- i = 0: nums[0] and nums[1] are not equal, so we skip this operation.
+- i = 1: nums[1] and nums[2] are equal, we multiply nums[1] by 2 and change nums[2] to 0. The array becomes [1,4,0,1,1,0].
+- i = 2: nums[2] and nums[3] are not equal, so we skip this operation.
+- i = 3: nums[3] and nums[4] are equal, we multiply nums[3] by 2 and change nums[4] to 0. The array becomes [1,4,0,2,0,0].
+- i = 4: nums[4] and nums[5] are equal, we multiply nums[4] by 2 and change nums[5] to 0. The array becomes [1,4,0,2,0,0].
+After that, we shift the 0's to the end, which gives the array [1,4,2,0,0,0].
+
+```
+
+### Example 2
+```
+Input: nums = [0,1]
+Output: [1,0]
+Explanation: No operation can be applied, we just shift the 0 to the end.
+
+```
+### Constraints
+
+- `2 <= nums.length <= 2000`
+- `0 <= nums[i] <= 1000`
+
+## Solution of Given Problem
+
+### Intuition and Approach
+
+The problem can be solved using a brute force approach or an optimized Technique.
+
+<Tabs>
+<tabItem value="Brute Force" label="Brute Force">
+
+### Approach 1:Brute Force (Naive)
+
+
+Brute Force Approach: 
+Iterate through the array and for each i, check if nums[i] is equal to nums[i + 1].
+If they are equal, multiply nums[i] by 2 and set nums[i + 1] to 0.
+After all operations are performed, create a new array by collecting all non-zero elements followed by all zero elements.
+#### Codes in Different Languages
+
+<Tabs>
+<TabItem value="C++" label="C++" default>
+<SolutionAuthor name="@AmruthaPariprolu"/>
+
+```cpp
+#include <iostream>
+#include <vector>
+
+std::vector<int> applyOperations(std::vector<int>& nums) {
+    int n = nums.size();
+    
+    for (int i = 0; i < n - 1; ++i) {
+        if (nums[i] == nums[i + 1]) {
+            nums[i] *= 2;
+            nums[i + 1] = 0;
+        }
+    }
+
+    std::vector<int> result;
+    for (int num : nums) {
+        if (num != 0) {
+            result.push_back(num);
+        }
+    }
+    while (result.size() < n) {
+        result.push_back(0);
+    }
+
+    return result;
+}
+
+int main() {
+    std::vector<int> nums = {1, 2, 2, 1, 1, 0};
+    std::vector<int> result = applyOperations(nums);
+    for (int num : result) {
+        std::cout << num << " ";
+    }
+    std::cout << std::endl;  // Output: 1 4 2 0 0 0
+    return 0;
+}
+
+```
+</TabItem>
+<TabItem value="Java" label="Java">
+<SolutionAuthor name="@AmruthaPariprolu"/>
+
+```java
+import java.util.*;
+
+public class ApplyOperations {
+    public static int[] applyOperations(int[] nums) {
+        int n = nums.length;
+        
+        for (int i = 0; i < n - 1; i++) {
+            if (nums[i] == nums[i + 1]) {
+                nums[i] *= 2;
+                nums[i + 1] = 0;
+            }
+        }
+        
+        int[] result = new int[n];
+        int idx = 0;
+        for (int num : nums) {
+            if (num != 0) {
+                result[idx++] = num;
+            }
+        }
+        
+        return result;
+    }
+
+    public static void main(String[] args) {
+        int[] nums = {1, 2, 2, 1, 1, 0};
+        int[] result = applyOperations(nums);
+        System.out.println(Arrays.toString(result));  // Output: [1, 4, 2, 0, 0, 0]
+    }
+}
+
+```
+
+
+</TabItem>
+<TabItem value="Python" label="Python">
+<SolutionAuthor name="@AmruthaPariprolu"/>
+
+```python
+def apply_operations(nums):
+    n = len(nums)
+    
+    for i in range(n - 1):
+        if nums[i] == nums[i + 1]:
+            nums[i] *= 2
+            nums[i + 1] = 0
+    
+    result = [num for num in nums if num != 0]
+    result.extend([0] * (n - len(result)))
+    
+    return result
+
+nums = [1, 2, 2, 1, 1, 0]
+print(apply_operations(nums))  # Output: [1, 4, 2, 0, 0, 0]
+
+```
+
+</TabItem>
+</Tabs>
+
+
+### Complexity Analysis
+
+- Time Complexity: $O(n)$
+-  for the operations and for shifting zeros.
+- Space Complexity: $O(n)$
+-   for the additional array.
+</tabItem>
+<tabItem value="Optimized approach" label="Optimized approach">
+
+### Approach 2: Optimized approach
+
+Optimized Approach: 
+Iterate through the array and perform the operations in place.
+After performing all operations, use two pointers to move all non-zero elements to the front and fill the remaining elements with zeros.
+#### Code in Different Languages
+
+<Tabs>
+<TabItem value="C++" label="C++" default>
+<SolutionAuthor name="@AmruthaPariprolu"/>
+
+```cpp
+#include <iostream>
+#include <vector>
+
+std::vector<int> applyOperations(std::vector<int>& nums) {
+    int n = nums.size();
+    
+    for (int i = 0; i < n - 1; ++i) {
+        if (nums[i] == nums[i + 1]) {
+            nums[i] *= 2;
+            nums[i + 1] = 0;
+        }
+    }
+
+    int index = 0;
+    for (int i = 0; i < n; ++i) {
+        if (nums[i] != 0) {
+            nums[index++] = nums[i];
+        }
+    }
+
+    while (index < n) {
+        nums[index++] = 0;
+    }
+
+    return nums;
+}
+
+int main() {
+    std::vector<int> nums = {1, 2, 2, 1, 1, 0};
+    std::vector<int> result = applyOperations(nums);
+    for (int num : result) {
+        std::cout << num << " ";
+    }
+    std::cout << std::endl;  // Output: 1 4 2 0 0 0
+    return 0;
+}
+
+```
+</TabItem>
+<TabItem value="Java" label="Java">
+<SolutionAuthor name="@AmruthaPariprolu"/>
+
+```java
+import java.util.*;
+
+public class ApplyOperations {
+    public static int[] applyOperations(int[] nums) {
+        int n = nums.length;
+        
+        for (int i = 0; i < n - 1; i++) {
+            if (nums[i] == nums[i + 1]) {
+                nums[i] *= 2;
+                nums[i + 1] = 0;
+            }
+        }
+        
+        int index = 0;
+        for (int i = 0; i < n; i++) {
+            if (nums[i] != 0) {
+                nums[index++] = nums[i];
+            }
+        }
+        
+        while (index < n) {
+            nums[index++] = 0;
+        }
+        
+        return nums;
+    }
+
+    public static void main(String[] args) {
+        int[] nums = {1, 2, 2, 1, 1, 0};
+        int[] result = applyOperations(nums);
+        System.out.println(Arrays.toString(result));  // Output: [1, 4, 2, 0, 0, 0]
+    }
+}
+
+```
+
+
+</TabItem>
+<TabItem value="Python" label="Python">
+<SolutionAuthor name="@AmruthaPariprolu"/>
+
+```python
+def apply_operations(nums):
+    n = len(nums)
+    
+    for i in range(n - 1):
+        if nums[i] == nums[i + 1]:
+            nums[i] *= 2
+            nums[i + 1] = 0
+    
+    index = 0
+    for i in range(n):
+        if nums[i] != 0:
+            nums[index] = nums[i]
+            index += 1
+    
+    while index < n:
+        nums[index] = 0
+        index += 1
+    
+    return nums
+
+nums = [1, 2, 2, 1, 1, 0]
+print(apply_operations(nums))  # Output: [1, 4, 2, 0, 0, 0]
+
+
+```
+
+</TabItem>
+</Tabs>
+
+#### Complexity Analysis
+
+- Time Complexity: $O(n)$
+-  for both operations and shifting.
+- Space Complexity: $O(1)$
+-  as no extra space is required.
+</tabItem>
+</Tabs>
+
+
+## Video Explanation of Given Problem
+
+    <LiteYouTubeEmbed
+      id="GLHChPK8VQ8"
+      params="autoplay=1&autohide=1&showinfo=0&rel=0"
+      title="Problem Explanation | Solution | Approach"
+      poster="maxresdefault"
+      webp 
+    />
+
+---
+
+<h2>Authors:</h2>
+
+<div style={{display: 'flex', flexWrap: 'wrap', justifyContent: 'space-between', gap: '10px'}}>
+{['AmruthaPariprolu'].map(username => (
+ <Author key={username} username={username} />
+))}
+</div>
+