Given an array arr[] of size N and a 2D array Q[][], consisting of queries of the following two types:
- 1 X Val: Update arr[X] = Val.
- 2 L R: Find the sum of array elements with alternating signs in the range [L, R].
Examples:
Input: arr[] = { 1, 2, 3, 4 }, Q[][] = { { 2, 0, 3 }, { 1, 1, 5 }, { 2, 1, 2 } }
Output:-2 2
Explanation:
Query1: Print (arr[0] – arr[1] + arr[2] – arr[3]) = 1 – 2 + 3 – 4 = -2
Query2: Updating arr[1] to 5 modifies arr[] to { 1, 5, 3, 4 }
Query3: Print (arr[1] – arr[2]) = 5 – 3 = 2Input: arr[] = { 4, 2, 6, 1, 8, 9, 2}, Q = { { 2, 1, 4 }, { 1, 3, 4 }, { 2, 0, 3 } }
Output: -11 5
Approach: The problem can be solved using Segment Tree. The idea is to traverse the array and check if the index of the array element is negative then multiply the array elements by -1. Follow the steps below to solve the problem:
- Traverse the array arr[] using variable i and check if the index i is odd or not. If found to be true, then update arr[i] = -Val.
- For query of type 1, X, Val, check if X is even or not. If found to be true, then update arr[X] = Val using segment tree.
- Otherwise, update arr[X] = -Val using segment tree.
- For query of type 2 L R:, check if L is even or not. If found to be true, then print the sum of array elements in the range [L, R] using the segment tree.
- Otherwise, find the sum of array elements in the range [L, R] using segment tree and print the obtained sum multiplying by -1.
Below is the implementation of the above approach :
C++
// C++ program to implement// the above approach#include <bits/stdc++.h>using namespace std;// Function to build the segment treevoid build(int tree[], int arr[], int start, int end, int index){ // If current node is a leaf node // of the segment tree if (start == end) { if (start % 2 == 0) { // Update tree[index] tree[index] = arr[start]; } else { // Update tree[index] tree[index] = -arr[start]; } return; } // Divide the segment tree int mid = start + (end - start) / 2; // Update on L segment tree build(tree, arr, start, mid, 2 * index + 1); // Update on R segment tree build(tree, arr, mid + 1, end, 2 * index + 2); // Find the sum from L subtree // and R subtree tree[index] = tree[2 * index + 1] + tree[2 * index + 2];}// Function to update elements at index pos// by val in the segment treevoid update(int tree[], int index, int start, int end, int pos, int val){ // If current node is a leaf node if (start == end) { // If current index is even if (start % 2 == 0) { // Update tree[index] tree[index] = val; } else { // Update tree[index] tree[index] = -val; } return; } // Divide the segment tree elements // into L and R subtree int mid = start + (end - start) / 2; // If element lies in L subtree if (mid >= pos) { update(tree, 2 * index + 1, start, mid, pos, val); } else { update(tree, 2 * index + 2, mid + 1, end, pos, val); } // Update tree[index] tree[index] = tree[2 * index + 1] + tree[2 * index + 2];}// Function to find the sum of array elements// in the range [L, R]int FindSum(int tree[], int start, int end, int L, int R, int index){ // If start and end not lies in // the range [L, R] if (L > end || R < start) { return 0; } // If start and end comleately lies // in the range [L, R] if (L <= start && R >= end) { return tree[index]; } int mid = start + (end - start) / 2; // Stores sum from left subtree int X = FindSum(tree, start, mid, L, R, 2 * index + 1); // Stores sum from right subtree int Y = FindSum(tree, mid + 1, end, L, R, 2 * index + 2); return X + Y;}int main(){ int arr[] = { 1, 2, 3, 4 }; int N = sizeof(arr) / sizeof(arr[0]); int tree[4 * N + 5] = { 0 }; build(tree, arr, 0, N - 1, 0); int Q[][3] = { { 2, 0, 3 }, { 1, 1, 5 }, { 2, 1, 2 } }; int cntQuey = 3; for (int i = 0; i < cntQuey; i++) { if (Q[i][0] == 1) { update(tree, 0, 0, N - 1, Q[i][1], Q[i][2]); } else { if (Q[i][1] % 2 == 0) { cout << FindSum(tree, 0, N - 1, Q[i][1], Q[i][2], 0) << " "; } else { cout << -FindSum(tree, 0, N - 1, Q[i][1], Q[i][2], 0) << " "; } } }} |
Java
// Java program to implement// the above approachimport java.util.*; class GFG{ // Function to build the segment treestatic void build(int tree[], int arr[], int start, int end, int index){ // If current node is a leaf node // of the segment tree if (start == end) { if (start % 2 == 0) { // Update tree[index] tree[index] = arr[start]; } else { // Update tree[index] tree[index] = -arr[start]; } return; } // Divide the segment tree int mid = start + (end - start) / 2; // Update on L segment tree build(tree, arr, start, mid, 2 * index + 1); // Update on R segment tree build(tree, arr, mid + 1, end, 2 * index + 2); // Find the sum from L subtree // and R subtree tree[index] = tree[2 * index + 1] + tree[2 * index + 2];}// Function to update elements at index pos// by val in the segment treestatic void update(int tree[], int index, int start, int end, int pos, int val){ // If current node is a leaf node if (start == end) { // If current index is even if (start % 2 == 0) { // Update tree[index] tree[index] = val; } else { // Update tree[index] tree[index] = -val; } return; } // Divide the segment tree elements // into L and R subtree int mid = start + (end - start) / 2; // If element lies in L subtree if (mid >= pos) { update(tree, 2 * index + 1, start, mid, pos, val); } else { update(tree, 2 * index + 2, mid + 1, end, pos, val); } // Update tree[index] tree[index] = tree[2 * index + 1] + tree[2 * index + 2];}// Function to find the sum of array elements// in the range [L, R]static int FindSum(int tree[], int start, int end, int L, int R, int index){ // If start and end not lies in // the range [L, R] if (L > end || R < start) { return 0; } // If start and end comleately lies // in the range [L, R] if (L <= start && R >= end) { return tree[index]; } int mid = start + (end - start) / 2; // Stores sum from left subtree int X = FindSum(tree, start, mid, L, R, 2 * index + 1); // Stores sum from right subtree int Y = FindSum(tree, mid + 1, end, L, R, 2 * index + 2); return X + Y;} // Driver Codepublic static void main(String[] args){ int arr[] = { 1, 2, 3, 4 }; int N = arr.length; int tree[] = new int[4 * N + 5]; Arrays.fill(tree, 0); build(tree, arr, 0, N - 1, 0); int Q[][] = { { 2, 0, 3 }, { 1, 1, 5 }, { 2, 1, 2 } }; int cntQuey = 3; for (int i = 0; i < cntQuey; i++) { if (Q[i][0] == 1) { update(tree, 0, 0, N - 1, Q[i][1], Q[i][2]); } else { if (Q[i][1] % 2 == 0) { System.out.print(FindSum(tree, 0, N - 1, Q[i][1], Q[i][2], 0) + " "); } else { System.out.print(-FindSum(tree, 0, N - 1, Q[i][1], Q[i][2], 0)+ " "); } } }}}// This code is contributed by code_hunt. |
Python3
# Python3 program to implement# the above approach# Function to build the segment treedef build(tree, arr, start, end, index): # If current node is a leaf node # of the segment tree if (start == end): if (start % 2 == 0): # Update tree[index] tree[index] = arr[start] else: # Update tree[index] tree[index] = -arr[start] return # Divide the segment tree mid = start + (end - start) // 2 # Update on L segment tree build(tree, arr, start, mid, 2 * index + 1) # Update on R segment tree build(tree, arr, mid + 1, end, 2 * index + 2) # Find the sum from L subtree # and R subtree tree[index] = tree[2 * index + 1] + tree[2 * index + 2]# Function to update elements at index pos# by val in the segment treedef update(tree, index, start, end, pos, val): # If current node is a leaf node if (start == end): # If current index is even if (start % 2 == 0): # Update tree[index] tree[index] = val else: # Update tree[index] tree[index] = -val return # Divide the segment tree elements # into L and R subtree mid = start + (end - start) // 2 # If element lies in L subtree if (mid >= pos): update(tree, 2 * index + 1, start, mid, pos, val) else: update(tree, 2 * index + 2, mid + 1, end, pos, val) # Update tree[index] tree[index] = tree[2 * index + 1] + tree[2 * index + 2]# Function to find the sum of array elements# in the range [L, R]def FindSum(tree, start, end, L, R, index): # If start and end not lies in # the range [L, R] if (L > end or R < start): return 0 #If start and end comleately lies #in the range [L, R] if (L <= start and R >= end): return tree[index] mid = start + (end - start) // 2 # Stores sum from left subtree X = FindSum(tree, start, mid, L, R, 2 * index + 1) # Stores sum from right subtree Y = FindSum(tree, mid + 1, end, L, R, 2 * index + 2) return X + Y # Driver codeif __name__ == '__main__': arr = [1, 2, 3, 4] N = len(arr) tree = [0 for i in range(4 * N + 5)] build(tree, arr, 0, N - 1, 0) Q = [ [ 2, 0, 3 ], [ 1, 1, 5 ], [ 2, 1, 2 ] ] cntQuey = 3 for i in range(cntQuey): if (Q[i][0] == 1): update(tree, 0, 0, N - 1, Q[i][1], Q[i][2]) else: if (Q[i][1] % 2 == 0): print(FindSum(tree, 0, N - 1, Q[i][1], Q[i][2], 0),end=" ") else: print(-FindSum(tree, 0, N - 1, Q[i][1], Q[i][2], 0),end=" ") # This code is contributed by mohit kumar 29 |
C#
// C# program to implement// the above approachusing System;class GFG { // Function to build the segment tree static void build(int[] tree, int[] arr, int start, int end, int index) { // If current node is a leaf node // of the segment tree if (start == end) { if (start % 2 == 0) { // Update tree[index] tree[index] = arr[start]; } else { // Update tree[index] tree[index] = -arr[start]; } return; } // Divide the segment tree int mid = start + (end - start) / 2; // Update on L segment tree build(tree, arr, start, mid, 2 * index + 1); // Update on R segment tree build(tree, arr, mid + 1, end, 2 * index + 2); // Find the sum from L subtree // and R subtree tree[index] = tree[2 * index + 1] + tree[2 * index + 2]; } // Function to update elements at index pos // by val in the segment tree static void update(int[] tree, int index, int start, int end, int pos, int val) { // If current node is a leaf node if (start == end) { // If current index is even if (start % 2 == 0) { // Update tree[index] tree[index] = val; } else { // Update tree[index] tree[index] = -val; } return; } // Divide the segment tree elements // into L and R subtree int mid = start + (end - start) / 2; // If element lies in L subtree if (mid >= pos) { update(tree, 2 * index + 1, start, mid, pos, val); } else { update(tree, 2 * index + 2, mid + 1, end, pos, val); } // Update tree[index] tree[index] = tree[2 * index + 1] + tree[2 * index + 2]; } // Function to find the sum of array elements // in the range [L, R] static int FindSum(int[] tree, int start, int end, int L, int R, int index) { // If start and end not lies in // the range [L, R] if (L > end || R < start) { return 0; } // If start and end comleately lies // in the range [L, R] if (L <= start && R >= end) { return tree[index]; } int mid = start + (end - start) / 2; // Stores sum from left subtree int X = FindSum(tree, start, mid, L, R, 2 * index + 1); // Stores sum from right subtree int Y = FindSum(tree, mid + 1, end, L, R, 2 * index + 2); return X + Y; } // Driver code static void Main() { int[] arr = { 1, 2, 3, 4 }; int N = arr.Length; int[] tree = new int[4 * N + 5]; build(tree, arr, 0, N - 1, 0); int[,] Q = { { 2, 0, 3 }, { 1, 1, 5 }, { 2, 1, 2 } }; int cntQuey = 3; for (int i = 0; i < cntQuey; i++) { if (Q[i, 0] == 1) { update(tree, 0, 0, N - 1, Q[i, 1], Q[i, 2]); } else { if (Q[i, 1] % 2 == 0) { Console.Write(FindSum(tree, 0, N - 1, Q[i, 1], Q[i, 2], 0) + " "); } else { Console.Write(-FindSum(tree, 0, N - 1, Q[i, 1], Q[i, 2], 0) + " "); } } } }}// This code is contributed by divyesh072019. |
Javascript
<script>// Javascript program to implement// the above approach// Function to build the segment treefunction build(tree, arr, start, end, index){ // If current node is a leaf node // of the segment tree if (start == end) { if (start % 2 == 0) { // Update tree[index] tree[index] = arr[start]; } else { // Update tree[index] tree[index] = -arr[start]; } return; } // Divide the segment tree var mid = start + parseInt((end - start) / 2); // Update on L segment tree build(tree, arr, start, mid, 2 * index + 1); // Update on R segment tree build(tree, arr, mid + 1, end, 2 * index + 2); // Find the sum from L subtree // and R subtree tree[index] = tree[2 * index + 1] + tree[2 * index + 2];}// Function to update elements at index pos// by val in the segment treefunction update(tree, index, start, end, pos, val){ // If current node is a leaf node if (start == end) { // If current index is even if (start % 2 == 0) { // Update tree[index] tree[index] = val; } else { // Update tree[index] tree[index] = -val; } return; } // Divide the segment tree elements // into L and R subtree var mid = start + parseInt((end - start) / 2); // If element lies in L subtree if (mid >= pos) { update(tree, 2 * index + 1, start, mid, pos, val); } else { update(tree, 2 * index + 2, mid + 1, end, pos, val); } // Update tree[index] tree[index] = tree[2 * index + 1] + tree[2 * index + 2];}// Function to find the sum of array elements// in the range [L, R]function FindSum(tree, start, end, L, R, index){ // If start and end not lies in // the range [L, R] if (L > end || R < start) { return 0; } // If start and end comleately lies // in the range [L, R] if (L <= start && R >= end) { return tree[index]; } var mid = start + parseInt((end - start) / 2); // Stores sum from left subtree var X = FindSum(tree, start, mid, L, R, 2 * index + 1); // Stores sum from right subtree var Y = FindSum(tree, mid + 1, end, L, R, 2 * index + 2); return X + Y;}var arr = [1, 2, 3, 4 ];var N = arr.length;var tree = Array(4*N+5).fill(0);build(tree, arr, 0, N - 1, 0);var Q = [ [ 2, 0, 3 ], [ 1, 1, 5 ], [ 2, 1, 2 ] ];var cntQuey = 3;for (var i = 0; i < cntQuey; i++) { if (Q[i][0] == 1) { update(tree, 0, 0, N - 1, Q[i][1], Q[i][2]); } else { if (Q[i][1] % 2 == 0) { document.write(FindSum(tree, 0, N - 1, Q[i][1], Q[i][2], 0) + " "); } else { document.write( -FindSum(tree, 0, N - 1, Q[i][1], Q[i][2], 0) + " "); } }}</script> |
Output:
-2 2
Time Complexity: O(|Q| * log(N))
Auxiliary Space: O(N)
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