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

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.

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).

#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

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

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.

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