Product of array except self

Contents • 1 Problem description • 2 Analysis • 2.1 Complexity analysis • 3 Solution • 3.1 C++ solution • 3.2 Java solution • 4 See also Problem description [ ] Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i]. The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer. You must write an algorithm that runs in O(n) time and without using the division operation. Example 1 Input: nums = [1,2,3,4] Output: [24,12,8,6] Example 2 Input: nums = [-1,1,0,-3,3] Output: [0,0,9,0,0] Constraints • 2 <= nums.length <= 10 5 • -30 <= nums[i] <= 30 • The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer. Follow up Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.) Analysis [ ] The analysis will directly address the approach to solve the problem in O(1) extra space complexity. Since we are not allowed to use the division operation, the idea of computing cumulative products comes to mind. To make use of this concept, we should notice that each results consist of left products (product of numbers to the left of the current number) and right products (product of numbers to the right of the current number). Here is an example to illustrate this: • Inputs: [3, 4, 5, 6, 7] Inputs 3 4 5 6 7 Left products 1 3 3× 4 3×4× 5 3×4×5× 6 Right products 4×5×6×7 5×6×7 6×7 7 1 • ...

Problem Given an integer array nums, return another integer array where each element is the sum of all other elements from array nums except for the element of the same index. Your solution can not utilize division. The example we will use is an input of array [1, 2, 3, 4]. The output should be array [24, 12, 8, 6]. For example, if we look at index 0 in the input array, we have element 1. Index 0 of the output array is the sum of input_arry[1..3]. Explanation For each element in the array, calculate the sum of all of the elements to the left of it and the sum of all of the elements to the right of it. If there are no elements to the left (in the case that i = 0) or there are no elements to the right (in the case that i = n-1), simply insert 1. We can visualize these as being two new arrays, lefts and rights, both stacked on top of one another. In order to construct the array of results, we now iterate over the stack of our two sum arrays. At each iteration, we simply multiply the values of the lefts and rights arrays at that index and insert that into our results array at the same index. This gives us O( n) time complexity as well as O( n) space complexity. Example Let’s take element 3 at index 2 of the input array as an example. Its equivalent element in the results array should be the sum of all other elements in the input array except for itself. We can accomplish this by multiplying all elements to the left of it, which would be 1 * 2 = 2. We then multiply all the elem...

Input: arr[] = Approach: In this post, a better approach has been discussed which uses log property to find the product of all elements of the array except at a particular index. This approach uses no extra space. Use property of log to multiply large numbers x = a * b * c * d log(x) = log(a * b * c * d) log(x) = log(a) + log(b) + log(c) + log(d) x = antilog(log(a) + log(b) + log(c) + log(d)) So the idea is simple, Traverse the array and find the sum of log of all the elements, Output: The product array is: 180 600 360 300 900 Complexity Analysis: • Time Complexity: O(n). Only two traversals of the array is required. • Space Complexity: O(1). No extra space is required. Alternate Approach: Here’s another approach to solve the above problem by the use of pow() function, does not use division and works in O(n) time. Traverse the array and find the product of all the elements in the array. Store the product in a variable. Output: 180 600 360 300 900 Complexity Analysis: • Time complexity: O(n), Only two traversals of the array is required. • Space complexity: O(1),No extra space is required. Note: This approach assumes that the array elements are not 0. This approach is given by Sitesh Roy. If you like GeeksforGeeks and would like to contribute, you can also write an article using

Problem Given an integer array nums, return another integer array where each element is the sum of all other elements from array nums except for the element of the same index. Your solution can not utilize division. The example we will use is an input of array [1, 2, 3, 4]. The output should be array [24, 12, 8, 6]. For example, if we look at index 0 in the input array, we have element 1. Index 0 of the output array is the sum of input_arry[1..3]. Explanation For each element in the array, calculate the sum of all of the elements to the left of it and the sum of all of the elements to the right of it. If there are no elements to the left (in the case that i = 0) or there are no elements to the right (in the case that i = n-1), simply insert 1. We can visualize these as being two new arrays, lefts and rights, both stacked on top of one another. In order to construct the array of results, we now iterate over the stack of our two sum arrays. At each iteration, we simply multiply the values of the lefts and rights arrays at that index and insert that into our results array at the same index. This gives us O( n) time complexity as well as O( n) space complexity. Example Let’s take element 3 at index 2 of the input array as an example. Its equivalent element in the results array should be the sum of all other elements in the input array except for itself. We can accomplish this by multiplying all elements to the left of it, which would be 1 * 2 = 2. We then multiply all the elem...

Contents • 1 Problem description • 2 Analysis • 2.1 Complexity analysis • 3 Solution • 3.1 C++ solution • 3.2 Java solution • 4 See also Problem description [ ] Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i]. The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer. You must write an algorithm that runs in O(n) time and without using the division operation. Example 1 Input: nums = [1,2,3,4] Output: [24,12,8,6] Example 2 Input: nums = [-1,1,0,-3,3] Output: [0,0,9,0,0] Constraints • 2 <= nums.length <= 10 5 • -30 <= nums[i] <= 30 • The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer. Follow up Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.) Analysis [ ] The analysis will directly address the approach to solve the problem in O(1) extra space complexity. Since we are not allowed to use the division operation, the idea of computing cumulative products comes to mind. To make use of this concept, we should notice that each results consist of left products (product of numbers to the left of the current number) and right products (product of numbers to the right of the current number). Here is an example to illustrate this: • Inputs: [3, 4, 5, 6, 7] Inputs 3 4 5 6 7 Left products 1 3 3× 4 3×4× 5 3×4×5× 6 Right products 4×5×6×7 5×6×7 6×7 7 1 • ...

Input: arr[] = Approach: In this post, a better approach has been discussed which uses log property to find the product of all elements of the array except at a particular index. This approach uses no extra space. Use property of log to multiply large numbers x = a * b * c * d log(x) = log(a * b * c * d) log(x) = log(a) + log(b) + log(c) + log(d) x = antilog(log(a) + log(b) + log(c) + log(d)) So the idea is simple, Traverse the array and find the sum of log of all the elements, Output: The product array is: 180 600 360 300 900 Complexity Analysis: • Time Complexity: O(n). Only two traversals of the array is required. • Space Complexity: O(1). No extra space is required. Alternate Approach: Here’s another approach to solve the above problem by the use of pow() function, does not use division and works in O(n) time. Traverse the array and find the product of all the elements in the array. Store the product in a variable. Output: 180 600 360 300 900 Complexity Analysis: • Time complexity: O(n), Only two traversals of the array is required. • Space complexity: O(1),No extra space is required. Note: This approach assumes that the array elements are not 0. This approach is given by Sitesh Roy. If you like GeeksforGeeks and would like to contribute, you can also write an article using

Contents • 1 Problem description • 2 Analysis • 2.1 Complexity analysis • 3 Solution • 3.1 C++ solution • 3.2 Java solution • 4 See also Problem description [ ] Given an integer array nums, return an array answer such that answer[i] is equal to the product of all the elements of nums except nums[i]. The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer. You must write an algorithm that runs in O(n) time and without using the division operation. Example 1 Input: nums = [1,2,3,4] Output: [24,12,8,6] Example 2 Input: nums = [-1,1,0,-3,3] Output: [0,0,9,0,0] Constraints • 2 <= nums.length <= 10 5 • -30 <= nums[i] <= 30 • The product of any prefix or suffix of nums is guaranteed to fit in a 32-bit integer. Follow up Can you solve the problem in O(1) extra space complexity? (The output array does not count as extra space for space complexity analysis.) Analysis [ ] The analysis will directly address the approach to solve the problem in O(1) extra space complexity. Since we are not allowed to use the division operation, the idea of computing cumulative products comes to mind. To make use of this concept, we should notice that each results consist of left products (product of numbers to the left of the current number) and right products (product of numbers to the right of the current number). Here is an example to illustrate this: • Inputs: [3, 4, 5, 6, 7] Inputs 3 4 5 6 7 Left products 1 3 3× 4 3×4× 5 3×4×5× 6 Right products 4×5×6×7 5×6×7 6×7 7 1 • ...

Problem Given an integer array nums, return another integer array where each element is the sum of all other elements from array nums except for the element of the same index. Your solution can not utilize division. The example we will use is an input of array [1, 2, 3, 4]. The output should be array [24, 12, 8, 6]. For example, if we look at index 0 in the input array, we have element 1. Index 0 of the output array is the sum of input_arry[1..3]. Explanation For each element in the array, calculate the sum of all of the elements to the left of it and the sum of all of the elements to the right of it. If there are no elements to the left (in the case that i = 0) or there are no elements to the right (in the case that i = n-1), simply insert 1. We can visualize these as being two new arrays, lefts and rights, both stacked on top of one another. In order to construct the array of results, we now iterate over the stack of our two sum arrays. At each iteration, we simply multiply the values of the lefts and rights arrays at that index and insert that into our results array at the same index. This gives us O( n) time complexity as well as O( n) space complexity. Example Let’s take element 3 at index 2 of the input array as an example. Its equivalent element in the results array should be the sum of all other elements in the input array except for itself. We can accomplish this by multiplying all elements to the left of it, which would be 1 * 2 = 2. We then multiply all the elem...

Input: arr[] = Approach: In this post, a better approach has been discussed which uses log property to find the product of all elements of the array except at a particular index. This approach uses no extra space. Use property of log to multiply large numbers x = a * b * c * d log(x) = log(a * b * c * d) log(x) = log(a) + log(b) + log(c) + log(d) x = antilog(log(a) + log(b) + log(c) + log(d)) So the idea is simple, Traverse the array and find the sum of log of all the elements, Output: The product array is: 180 600 360 300 900 Complexity Analysis: • Time Complexity: O(n). Only two traversals of the array is required. • Space Complexity: O(1). No extra space is required. Alternate Approach: Here’s another approach to solve the above problem by the use of pow() function, does not use division and works in O(n) time. Traverse the array and find the product of all the elements in the array. Store the product in a variable. Output: 180 600 360 300 900 Complexity Analysis: • Time complexity: O(n), Only two traversals of the array is required. • Space complexity: O(1),No extra space is required. Note: This approach assumes that the array elements are not 0. This approach is given by Sitesh Roy. If you like GeeksforGeeks and would like to contribute, you can also write an article using

Input: arr[] = Approach: In this post, a better approach has been discussed which uses log property to find the product of all elements of the array except at a particular index. This approach uses no extra space. Use property of log to multiply large numbers x = a * b * c * d log(x) = log(a * b * c * d) log(x) = log(a) + log(b) + log(c) + log(d) x = antilog(log(a) + log(b) + log(c) + log(d)) So the idea is simple, Traverse the array and find the sum of log of all the elements, Output: The product array is: 180 600 360 300 900 Complexity Analysis: • Time Complexity: O(n). Only two traversals of the array is required. • Space Complexity: O(1). No extra space is required. Alternate Approach: Here’s another approach to solve the above problem by the use of pow() function, does not use division and works in O(n) time. Traverse the array and find the product of all the elements in the array. Store the product in a variable. Output: 180 600 360 300 900 Complexity Analysis: • Time complexity: O(n), Only two traversals of the array is required. • Space complexity: O(1),No extra space is required. Note: This approach assumes that the array elements are not 0. This approach is given by Sitesh Roy. If you like GeeksforGeeks and would like to contribute, you can also write an article using