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A prefix sum array is another array prefixSum[] of the same size, such that the value of prefixSum[i] is arr[0] + arr[1] + arr[2] . . . arr[i].\n\nExamples: \nInput: arr[] = {10, 20, 10, 5, 15}\nOutput: prefixSum[] = {10, 30, 40, 45, 60}\nExplanation: While traversing the array, update the element by adding it with its previous element.\nprefixSum[0] = 10, \nprefixSum[1] = prefixSum[0] + arr[1] = 30, \nprefixSum[2] = prefixSum[1] + arr[2] = 40 and so on.","datePublished":"2024-08-02T12:10:38Z","dateModified":"2024-08-02T12:10:38Z","author":{"@type":"Organization","name":"Leetcode with dani","url":"https://telemetr.io/fa/channels/1979744670-leetcodeq","image":"https://img.tlmtr.io/c/29YO1E/5832277572212345105?ty=x"},"publisher":{"@type":"Organization","name":"Leetcode with dani","url":"https://telemetr.io/fa/channels/1979744670-leetcodeq","image":"https://img.tlmtr.io/c/29YO1E/5832277572212345105?ty=x"},"commentCount":0,"interactionStatistic":[{"@type":"InteractionCounter","interactionType":"https://schema.org/ViewAction","userInteractionCount":296},{"@type":"InteractionCounter","interactionType":"https://schema.org/LikeAction","userInteractionCount":2}]}}]}

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def fillPrefixSum(arr, n, prefixSum):

  prefixSum[0] = arr[0]

  # Adding present element
  # with previous element
  for i in range(1, n):
    prefixSum[i] = prefixSum[i - 1] + arr[i]


# Driver code
if __name__ == '__main__':
arr = [10, 4, 16, 20]
n = len(arr)

# Function call
prefixSum = [0 for i in range(n + 1)]

fillPrefixSum(arr, n, prefixSum)

for i in range(n):
  print(prefixSum[i], " ", end="")

# This code is contributed
# by Anant Agarwal.

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#Q18 #leet_codeQ18 Easy 1480 . Given an array nums. We define a running sum of an array as runningSum[i] = sum(nums[0]…nums[i]). Return the running sum of nums. Example 1: Input: nums = [1,2,3,4] Output: [1,3,6,10] Explanation: Running sum is obtained as follows: [1, 1+2, 1+2+3, 1+2+3+4]. Example 2: Input: nums = [1,1,1,1,1] Output: [1,2,3,4,5] Explanation: Running sum is obtained as follows: [1, 1+1, 1+1+1, 1+1+1+1, 1+1+1+1+1]. Example 3: Input: nums = [3,1,2,10,1] Output: [3,4,6,16,17] Constraints: 1 <= nums.length <= 1000 -10^6 <= nums[i] <= 10^6

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Example 2: Input : Arr[] = {6, 7, 8, 10}, Q = 2 queries[] = {0, 3, 1, 2} Output : 7 7

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Example 1: Input : Arr[] = {1, 2, 3, 4, 5}, Q = 3 queries[] = {0, 2, 1, 3, 0, 4} Output : 2 3 3 Explanation: Here we can see that the array of integers is [1, 2, 3, 4, 5]. Query 1: L = 0 and R = 2 Sum = 6 Integer Count = 3 So, Mean is 2 Query 2: L = 1 and R = 3 Sum = 9 Integer Count = 3 So, Mean is 3 Query 3: L = 0 and R = 4 Sum = 15 Integer Count = 5 So, the Mean is 3. So, In the end, the function will return the array [2, 3, 3] as an answer.

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Mean of range in array #Q17 Geeks for Geeks Given an array of n integers and q queries. Write a program to find floor value of mean in range l to r for each query in a new line. Queries are given by an array queries[] of size 2*q. Here queries[2*i] denote l and queries[2*i+1] denote r for i-th query (0<= i <q).

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#leet_codeQ16 #Q_209 #Easy #Prefix_sum 1991. Find the Middle Index in Array Hint Given a 0-indexed integer array nums, find the leftmost middleIndex (i.e., the smallest amongst all the possible ones). A middleIndex is an index where nums[0] + nums[1] + ... + nums[middleIndex-1] == nums[middleIndex+1] + nums[middleIndex+2] + ... + nums[nums.length-1]. If middleIndex == 0, the left side sum is considered to be 0. Similarly, if middleIndex == nums.length - 1, the right side sum is considered to be 0. Return the leftmost middleIndex that satisfies the condition, or -1 if there is no such index. Example 1: Input: nums = [2,3,-1,8,4] Output: 3 Explanation: The sum of the numbers before index 3 is: 2 + 3 + -1 = 4 The sum of the numbers after index 3 is: 4 = 4 Example 2: Input: nums = [1,-1,4] Output: 2 Explanation: The sum of the numbers before index 2 is: 1 + -1 = 0 The sum of the numbers after index 2 is: 0 Example 3: Input: nums = [2,5] Output: -1 Explanation: There is no valid middleIndex. Constraints: 1 <= nums.length <= 100 -1000 <= nums[i] <= 1000

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1991. Find the Middle Index in Array Hint Given a 0-indexed integer array nums, find the leftmost middleIndex (i.e., the smallest amongst all the possible ones). A middleIndex is an index where nums[0] + nums[1] + ... + nums[middleIndex-1] == nums[middleIndex+1] + nums[middleIndex+2] + ... + nums[nums.length-1]. If middleIndex == 0, the left side sum is considered to be 0. Similarly, if middleIndex == nums.length - 1, the right side sum is considered to be 0. Return the leftmost middleIndex that satisfies the condition, or -1 if there is no such index. Example 1: Input: nums = [2,3,-1,8,4] Output: 3 Explanation: The sum of the numbers before index 3 is: 2 + 3 + -1 = 4 The sum of the numbers after index 3 is: 4 = 4 Example 2: Input: nums = [1,-1,4] Output: 2 Explanation: The sum of the numbers before index 2 is: 1 + -1 = 0 The sum of the numbers after index 2 is: 0 Example 3: Input: nums = [2,5] Output: -1 Explanation: There is no valid middleIndex. Constraints: 1 <= nums.length <= 100 -1000 <= nums[i] <= 1000

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Answer:

def prefix_sum(arr):
    prefix_sum_arr = [0] * len(arr)
    prefix_sum_arr[0] = arr[0]
    for i in range(1, len(arr)):
        prefix_sum_arr[i] = prefix_sum_arr[i - 1] + arr[i]
    return prefix_sum_arr

arr = [10, 20, 10, 5, 15]
result = prefix_sum(arr)
print(result)  # Output: [10, 30, 40, 45, 60]

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Prefix Sum Array – Implementation and Applications in Competitive Programming Given an array arr[] of size N, find the prefix sum of the array. A prefix sum array is another array prefixSum[] of the same size, such that the value of prefixSum[i] is arr[0] + arr[1] + arr[2] . . . arr[i]. Examples: Input: arr[] = {10, 20, 10, 5, 15} Output: prefixSum[] = {10, 30, 40, 45, 60} Explanation: While traversing the array, update the element by adding it with its previous element. prefixSum[0] = 10, prefixSum[1] = prefixSum[0] + arr[1] = 30, prefixSum[2] = prefixSum[1] + arr[2] = 40 and so on.