Temple Offerings

30 May, 2022
Consider a devotee wishing to give offerings to temples along with a mountain range. The temples are located in a row at different heights. Each temple should receive at least one offer. If two adjacent temples are at different altitudes, then the temple that is higher up should receive more offerings than the one that is lower down. If two adjacent temples are at the same height, then their offerings relative to each other do not matter. Given the number of temples and the heights of the temples in order, find the minimum number of offerings to bring.
Examples:
Input  : 3
         1 2 2
Output : 4
All temples must receive at-least one offering.
Now, the second temple is at a higher altitude
compared to the first one. Thus it receives one
extra offering. 
The second temple and third temple are at the 
same height, so we do not need to modify the 
offerings. Offerings given are therefore: 1, 2,
1 giving a total of 4.

Input  : 6
         1 4 3 6 2 1
Output : 10
We can distribute the offerings in the following
way, 1, 2, 1, 3, 2, 1. The second temple has to 
receive more offerings than the first due to its 
height being higher. The fourth must receive more
than the fifth, which in turn must receive more 
than the sixth. Thus the total becomes 10.
We notice that each temple can either be above, below or at the same level as the temple next to it. The offerings required at each temple are equal to the maximum length of the chain of temples at a lower height as shown in the image.
Naive Approach To follow the given rule, a temple must be offered at least x+1 where x is the maximum of the following two.
Number of temples on left in increasing order.Number of temples on right in increasing order.
Number of temples on left in increasing order.
Number of temples on right in increasing order.
A naive method of solving this problem would be for each temple, go to the left until altitude increases, and do the same for the right.
C++
Java
Python3
C#
PHP
Javascript
// Program to find minimum total offerings required#include <iostream>using namespace std; // Returns minimum offerings requiredint offeringNumber(int n, int templeHeight[]){    int sum = 0;  // Initialize result     // Go through all temples one by one    for (int i = 0; i < n; ++i)    {        // Go to left while height keeps increasing        int left = 0, right = 0;        for (int j = i - 1; j >= 0; --j)        {            if (templeHeight[j] < templeHeight[j + 1])                ++left;            else                break;        }         // Go to right while height keeps increasing        for (int j = i + 1; j < n; ++j)        {            if (templeHeight[j] < templeHeight[j - 1])                ++right;            else                break;        }         // This temple should offer maximum of two        // values to follow the rule.        sum += max(right, left) + 1;    }     return sum;} // Driver codeint main(){    int arr1[3] = {1, 2, 2};    cout << offeringNumber(3, arr1) << "\n";    int arr2[6] = {1, 4, 3, 6, 2, 1};    cout << offeringNumber(6, arr2) << "\n";    return 0;}
// Program to find minimum// total offerings requiredimport java.io.*; class GFG{     // Returns minimum// offerings requiredstatic int offeringNumber(int n,                          int templeHeight[]){    int sum = 0; // Initialize result     // Go through all    // temples one by one    for (int i = 0; i < n; ++i)    {        // Go to left while        // height keeps increasing        int left = 0, right = 0;        for (int j = i - 1; j >= 0; --j)        {            if (templeHeight[j] <                templeHeight[j + 1])                ++left;            else                break;        }         // Go to right while        // height keeps increasing        for (int j = i + 1; j < n; ++j)        {            if (templeHeight[j] <                templeHeight[j - 1])                ++right;            else                break;        }         // This temple should offer        // maximum of two values        // to follow the rule.        sum += Math.max(right, left) + 1;    }     return sum;} // Driver codepublic static void main (String[] args){int arr1[] = {1, 2, 2};System.out.println(offeringNumber(3, arr1));int arr2[] = {1, 4, 3,              6, 2, 1};System.out.println(offeringNumber(6, arr2));}} // This code is contributed by akt_mit
# Program to find minimum total# offerings required. # Returns minimum offerings requireddef offeringNumber(n, templeHeight):    sum = 0 # Initialize result         # Go through all temples one by one    for i in range(n):                 # Go to left while height        # keeps increasing        left = 0        right = 0        for j in range(i - 1, -1, -1):            if (templeHeight[j] < templeHeight[j + 1]):                left += 1            else:                break                         # Go to right while height        # keeps increasing        for j in range(i + 1, n):            if (templeHeight[j] < templeHeight[j - 1]):                right += 1            else:                break                         # This temple should offer maximum        # of two values to follow the rule.        sum += max(right, left) + 1    return sum # Driver Codearr1 = [1, 2, 2]print(offeringNumber(3, arr1))arr2 = [1, 4, 3, 6, 2, 1]print(offeringNumber(6, arr2)) # This code is contributed# by sahilshelangia
// Program to find minimum// total offerings requiredusing System; class GFG{     // Returns minimum// offerings requiredstatic int offeringNumber(int n,                          int []templeHeight){    int sum = 0; // Initialize result     // Go through all    // temples one by one    for (int i = 0; i < n; ++i)    {        // Go to left while        // height keeps increasing        int left = 0, right = 0;        for (int j = i - 1; j >= 0; --j)        {            if (templeHeight[j] <                templeHeight[j + 1])                ++left;            else                break;        }         // Go to right while        // height keeps increasing        for (int j = i + 1; j < n; ++j)        {            if (templeHeight[j] <                templeHeight[j - 1])                ++right;            else                break;        }         // This temple should offer        // maximum of two values        // to follow the rule.        sum += Math.Max(right, left) + 1;    }     return sum;} // Driver codestatic public void Main (){    int []arr1 = {1, 2, 2};    Console.WriteLine(offeringNumber(3, arr1));         int []arr2 = {1, 4, 3,                  6, 2, 1};    Console.WriteLine(offeringNumber(6, arr2));}} // This code is contributed by aj_36
<?php// Program to find minimum total offerings required // Returns minimum offerings requiredfunction offeringNumber($n, $templeHeight){    $sum = 0; // Initialize result     // Go through all temples one by one    for ($i = 0; $i < $n; ++$i)    {        // Go to left while height keeps increasing        $left = 0; $right = 0;        for ($j = $i - 1; $j >= 0; --$j)        {            if ($templeHeight[$j] < $templeHeight[$j + 1])                ++$left;            else                break;        }         // Go to right while height keeps increasing        for ($j = $i + 1; $j < $n; ++$j)        {            if ($templeHeight[$j] < $templeHeight[$j - 1])                ++$right;            else                break;        }         // This temple should offer maximum of two        // values to follow the rule.        $sum += max($right, $left) + 1;    }     return $sum;} // Driver code    $arr1 = array (1, 2, 2);    echo offeringNumber(3, $arr1) , "\n";    $arr2 = array (1, 4, 3, 6, 2, 1);    echo offeringNumber(6, $arr2) ,"\n";     // This code is contributed by ajit?>
<script>    // Program to find minimum    // total offerings required         // Returns minimum    // offerings required    function offeringNumber(n, templeHeight)    {        let sum = 0; // Initialize result         // Go through all        // temples one by one        for (let i = 0; i < n; ++i)        {            // Go to left while            // height keeps increasing            let left = 0, right = 0;            for (let j = i - 1; j >= 0; --j)            {                if (templeHeight[j] < templeHeight[j + 1])                    ++left;                else                    break;            }             // Go to right while            // height keeps increasing            for (let j = i + 1; j < n; ++j)            {                if (templeHeight[j] < templeHeight[j - 1])                    ++right;                else                    break;            }             // This temple should offer            // maximum of two values            // to follow the rule.            sum += Math.max(right, left) + 1;        }         return sum;    }               let arr1 = [1, 2, 2];    document.write(offeringNumber(3, arr1) + "</br>");          let arr2 = [1, 4, 3, 6, 2, 1];    document.write(offeringNumber(6, arr2));</script>
4
10
Time complexity: O(n2) Space complexity: O(1) 
Dynamic Programming Approach By using Dynamic Programming, we can improve the time complexity. In this method, we create a structure of length n which maintains the maximum decreasing chain to the left of each temple and the maximum decreasing chain to the right of each temple. We go through once from 0 to N setting the value of left for each temple. We then go from N to 0 setting the value of right for each temple. We then compare the two and pick the maximum for each temple.
C++
Java
Python3
C#
// C++ Program to find total offerings required#include <iostream>using namespace std; // To store count of increasing order temples// on left and right (including current temple)struct Temple {    int L;    int R;}; // Returns count of minimum offerings for// n temples of given heights.int offeringNumber(int n, int templeHeight[]){    // Initialize counts for all temples    Temple chainSize[n];    for (int i = 0; i < n; ++i) {        chainSize[i].L = -1;        chainSize[i].R = -1;    }     // Values corner temples    chainSize[0].L = 1;    chainSize[n - 1].R = 1;     // Filling left and right values using same    // values of previous(or next)    for (int i = 1; i < n; ++i) {        if (templeHeight[i - 1] < templeHeight[i])            chainSize[i].L = chainSize[i - 1].L + 1;        else            chainSize[i].L = 1;    }    for (int i = n - 2; i >= 0; --i) {        if (templeHeight[i + 1] < templeHeight[i])            chainSize[i].R = chainSize[i + 1].R + 1;        else            chainSize[i].R = 1;    }     // Computing max of left and right for all    // temples and returning sum.    int sum = 0;    for (int i = 0; i < n; ++i)        sum += max(chainSize[i].L, chainSize[i].R);    return sum;} // Driver functionint main(){    int arr1[3] = { 1, 2, 2 };    cout << offeringNumber(3, arr1) << "\n";    int arr2[6] = { 1, 4, 3, 6, 2, 1 };    cout << offeringNumber(6, arr2) << "\n";    return 0;}
// Java program to find total offerings requiredimport java.util.*; class GFG {     // To store count of increasing order temples    // on left and right (including current temple)    public static class Temple {        public int L;        public int R;    };     // Returns count of minimum offerings for    // n temples of given heights.    static int offeringNumber(int n, int[] templeHeight)    {         // Initialize counts for all temples        Temple[] chainSize = new Temple[n];         for (int i = 0; i < n; ++i) {            chainSize[i] = new Temple();            chainSize[i].L = -1;            chainSize[i].R = -1;        }         // Values corner temples        chainSize[0].L = 1;        chainSize[n - 1].R = 1;         // Filling left and right values        // using same values of        // previous(or next)        for (int i = 1; i < n; ++i) {            if (templeHeight[i - 1] < templeHeight[i])                chainSize[i].L = chainSize[i - 1].L + 1;            else                chainSize[i].L = 1;        }         for (int i = n - 2; i >= 0; --i) {            if (templeHeight[i + 1] < templeHeight[i])                chainSize[i].R = chainSize[i + 1].R + 1;            else                chainSize[i].R = 1;        }         // Computing max of left and right for all        // temples and returning sum.        int sum = 0;        for (int i = 0; i < n; ++i)            sum += Math.max(chainSize[i].L, chainSize[i].R);         return sum;    }     // Driver code    public static void main(String[] s)    {        int[] arr1 = { 1, 2, 2 };        System.out.println(offeringNumber(3, arr1));         int[] arr2 = { 1, 4, 3, 6, 2, 1 };        System.out.println(offeringNumber(6, arr2));    }} // This code is contributed by pratham76
# Python3 program to find temple# offerings requiredfrom typing import List # To store count of increasing order temples# on left and right (including current temple)  class Temple:    def __init__(self, l: int, r: int):         self.L = l        self.R = r # Returns count of minimum offerings for# n temples of given heights.  def offeringNumber(n: int,                   templeHeight: List[int]) -> int:     # Initialize counts for all temples    chainSize = [0] * n     for i in range(n):        chainSize[i] = Temple(-1, -1)     # Values corner temples    chainSize[0].L = 1    chainSize[-1].R = 1     # Filling left and right values    # using same values of previous(or next    for i in range(1, n):        if templeHeight[i - 1] < templeHeight[i]:            chainSize[i].L = chainSize[i - 1].L + 1        else:            chainSize[i].L = 1     for i in range(n - 2, -1, -1):        if templeHeight[i + 1] < templeHeight[i]:            chainSize[i].R = chainSize[i + 1].R + 1        else:            chainSize[i].R = 1     # Computing max of left and right for all    # temples and returning sum    sm = 0    for i in range(n):        sm += max(chainSize[i].L,                  chainSize[i].R)     return sm  # Driver codeif __name__ == '__main__':     arr1 = [1, 2, 2]    print(offeringNumber(3, arr1))     arr2 = [1, 4, 3, 6, 2, 1]    print(offeringNumber(6, arr2)) # This code is contributed by Rajat Srivastava
// C# program to find total offerings requiredusing System; class GFG {     // To store count of increasing order temples    // on left and right (including current temple)    public class Temple {        public int L;        public int R;    };     // Returns count of minimum offerings for    // n temples of given heights.    static int offeringNumber(int n, int[] templeHeight)    {         // Initialize counts for all temples        Temple[] chainSize = new Temple[n];         for (int i = 0; i < n; ++i) {            chainSize[i] = new Temple();            chainSize[i].L = -1;            chainSize[i].R = -1;        }         // Values corner temples        chainSize[0].L = 1;        chainSize[n - 1].R = 1;         // Filling left and right values        // using same values of        // previous(or next)        for (int i = 1; i < n; ++i) {            if (templeHeight[i - 1] < templeHeight[i])                chainSize[i].L = chainSize[i - 1].L + 1;            else                chainSize[i].L = 1;        }        for (int i = n - 2; i >= 0; --i) {            if (templeHeight[i + 1] < templeHeight[i])                chainSize[i].R = chainSize[i + 1].R + 1;            else                chainSize[i].R = 1;        }         // Computing max of left and right for all        // temples and returning sum.        int sum = 0;        for (int i = 0; i < n; ++i)            sum += Math.Max(chainSize[i].L, chainSize[i].R);         return sum;    }     // Driver code    static void Main()    {        int[] arr1 = { 1, 2, 2 };        Console.Write(offeringNumber(3, arr1) + "\n");         int[] arr2 = { 1, 4, 3, 6, 2, 1 };        Console.Write(offeringNumber(6, arr2) + "\n");    }} // This code is contributed by rutvik_56
4
10
Time complexity: O(n) Space complexity: O(n)
Greedy Approach:
If we somehow manage to make sure that the temple at higher mountain is getting more offerings than our problem is solved. For this we can make use of greedy (since we have to compare only the neighbors of current index). The approach is to do two traversals (in two directions), first one to make sure that the temple gets more offerings than the left temple (at higher position) and second one to make sure that the temple at higher position from the right gets more offerings. 
C++
Java
// C++ Program to find total offerings required#include <iostream>using namespace std; int templeOfferings(int nums[], int n){  // to find the total offerings in both directions  int offerings[n];   // start off by giving one offering to the first  // temple  offerings[0] = 1;   // to make sure that the temple at ith position gets  // more offerings if it is at a greater height than  // the left one  for (int i = 1; i < n; i++)  {    if (nums[i] > nums[i - 1])      offerings[i] = offerings[i - 1] + 1;    else      offerings[i] = 1;  }   // to make sure that the temple at ith position gets  // more offerings if it is at a greater height than  // the right one  for (int i = n - 2; i >= 0; i--)  {    if (nums[i] > nums[i + 1] && offerings[i] <= offerings[i + 1])      offerings[i] = offerings[i + 1] + 1;  }   // total offerings  int sum = 0;  for (int val : offerings)    sum += val;  return sum;} // Driver functionint main(){  int arr[] = { 1, 4, 3, 6, 2, 1 };  int n = sizeof(arr)/sizeof(arr[0]);  cout << (templeOfferings(arr, n));  return 0;} // This code is contributed by kothavvsaakash
/*package whatever //do not write package name here */ import java.io.*; class GFG {    public static void main(String[] args)    {        int[] arr = { 1, 4, 3, 6, 2, 1 };        int n = arr.length;        System.out.println(templeOfferings(arr, n));    }     private static int templeOfferings(int[] nums, int n)    {        // to find the total offerings in both directions        int[] offerings = new int[n];         // start off by giving one offering to the first        // temple        offerings[0] = 1;         // to make sure that the temple at ith position gets        // more offerings if it is at a greater height than        // the left one        for (int i = 1; i < n; i++) {            if (nums[i] > nums[i - 1])                offerings[i] = offerings[i - 1] + 1;            else                offerings[i] = 1;        }         // to make sure that the temple at ith position gets        // more offerings if it is at a greater height than        // the right one        for (int i = n - 2; i >= 0; i--) {            if (nums[i] > nums[i + 1]                && offerings[i] <= offerings[i + 1])                offerings[i] = offerings[i + 1] + 1;        }         // total offerings        int sum = 0;        for (int val : offerings)            sum += val;        return sum;    }}
10
Time complexity: O(n) 
Space complexity: O(n)
This article is contributed by Aditya Kamath and improved by Vishal Jha. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
jit_t
sahilshelangia
rajatsri94
rutvik_56
pratham76
mukesh07
sherlockvj
kashishsoda
sumitgumber28
rkbhola5
kothavvsaakash
Dynamic Programming
Dynamic Programming
Writing code in comment?
Please use ide.geeksforgeeks.org,
generate link and share the link here.