Queries to calculate sum by alternating signs of array elements in a given range

Last Updated : 12 May, 2021

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 = 2

Input: 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 tree
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
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]
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 approach
import java.util.*;
  
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
public 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 tree
def 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 tree
def 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 code
if __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 approach
using 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 tree
function 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 tree
function 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)

Queries to find the minimum index in a range [L, R] having at least value X with updates

single__loop

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Queries to calculate sum by alternating signs of array elements in a given range

C++

Java

Python3

C#

Javascript

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