Given an array arr[] of length N, the task is to find the minimum number of prizes required such that if two elements are adjacent, then elements with smaller value gets a less number of prizes compared to its adjacent elements with greater value.
Note: Each elements will get at least one prize.
Examples:
Input: arr[] = {1, 2, 2, 3}
Output: 6
Explanation:
Element at index {0} will get {1} prize.
Element at index {1} will get {2} prizes.
Element at index {2} will get {1} prizes.
Element at index {3} will get {2} prizes.
So, the total number of prizes required to satisfy
the above conditions are 6Input: arr[] = {3, 2, 2, 1}
Output: 6
Explanation:
Element at index {0} will get {2} prize.
Element at index {1} will get {1} prizes.
Element at index {2} will get {2} prizes.
Element at index {3} will get {1} prizes.
So, the total number of prizes required to satisfy
the above conditions are 6
Naive Approach: Traverse over the elements of the array and for each element of the array find the consecutive smaller elements at the left of the element and find the consecutive smaller elements on the right of that index.
Prize at index i = max(Consecutive smaller elements at left, Consecutive smaller elements at right, 1)
Below is the implementation of the above approach:
C++
// C++ implementation to find the// minimum prizes required such// that adjacent smaller elements// gets less number of prizes#include <bits/stdc++.h>using namespace std;// Function to find the minimum// number of required such that// adjacent smaller elements gets// less number of prizesint findMinPrizes(int arr[], int n){ int totalPrizes = 0, j, x, y; // Loop to iterate over every // elements of the array for (int i = 0; i < n; i++) { x = 1; j = i; // Loop to find the consecutive // smaller elements at left while (j > 0 && arr[j] > arr[j - 1]) { x++; j--; } j = i; y = 1; // Loop to find the consecutive // smaller elements at right while (j < n - 1 && arr[j] > arr[j + 1]) { y++; j++; } totalPrizes += max({ x, y }); } cout << totalPrizes << endl; return 0;}// Driver Codeint main(){ int arr[] = { 1, 2, 2, 3 }; int n = sizeof(arr) / sizeof(arr[0]); findMinPrizes(arr, n);} |
Java
// Java implementation to find the// minimum prizes required such// that adjacent smaller elements// gets less number of prizesimport java.util.*;class GFG{// Function to find the minimum// number of required such that// adjacent smaller elements gets// less number of prizesstatic int findMinPrizes(int arr[], int n){ int totalPrizes = 0, j, x, y; // Loop to iterate over every // elements of the array for (int i = 0; i < n; i++) { x = 1; j = i; // Loop to find the consecutive // smaller elements at left while (j > 0 && arr[j] > arr[j - 1]) { x++; j--; } j = i; y = 1; // Loop to find the consecutive // smaller elements at right while (j < n - 1 && arr[j] > arr[j + 1]) { y++; j++; } totalPrizes += Math.max(x, y ); } System.out.print(totalPrizes + "\n"); return 0;}// Driver Codepublic static void main(String[] args){ int arr[] = { 1, 2, 2, 3 }; int n = arr.length; findMinPrizes(arr, n);}}// This code is contributed by gauravrajput1 |
Python3
# Python3 implementation to find the # minimum prizes required such # that adjacent smaller elements # gets less number of prizes # Function to find the minimum # number of required such that # adjacent smaller elements gets # less number of prizes def findMinPrizes(arr, n): totalPrizes = 0 # Loop to iterate over every # elements of the array for i in range(n): x = 1 j = i # Loop to find the consecutive # smaller elements at left while (j > 0 and arr[j] > arr[j - 1]): x += 1 j -= 1 j = i y = 1 # Loop to find the consecutive # smaller elements at right while (j < n - 1 and arr[j] > arr[j + 1]): y += 1 j += 1 totalPrizes += max(x, y) print(totalPrizes)# Driver codearr = [ 1, 2, 2, 3 ]n = len(arr)findMinPrizes(arr, n)# This code is contributed by stutipathak31jan |
C#
// C# implementation to find the// minimum prizes required such// that adjacent smaller elements// gets less number of prizesusing System;class GFG{// Function to find the minimum// number of required such that// adjacent smaller elements gets// less number of prizesstatic int findMinPrizes(int []arr, int n){ int totalPrizes = 0, j, x, y; // Loop to iterate over every // elements of the array for(int i = 0; i < n; i++) { x = 1; j = i; // Loop to find the consecutive // smaller elements at left while (j > 0 && arr[j] > arr[j - 1]) { x++; j--; } j = i; y = 1; // Loop to find the consecutive // smaller elements at right while (j < n - 1 && arr[j] > arr[j + 1]) { y++; j++; } totalPrizes += Math.Max(x, y); } Console.Write(totalPrizes + "\n"); return 0;}// Driver Codepublic static void Main(String[] args){ int []arr = { 1, 2, 2, 3 }; int n = arr.Length; findMinPrizes(arr, n);}}// This code is contributed by Amit Katiyar |
Javascript
<script> // Javascript implementation to find the // minimum prizes required such // that adjacent smaller elements // gets less number of prizes // Function to find the minimum // number of required such that // adjacent smaller elements gets // less number of prizes function findMinPrizes(arr, n) { let totalPrizes = 0, j, x, y; // Loop to iterate over every // elements of the array for (let i = 0; i < n; i++) { x = 1; j = i; // Loop to find the consecutive // smaller elements at left while (j > 0 && arr[j] > arr[j - 1]) { x++; j--; } j = i; y = 1; // Loop to find the consecutive // smaller elements at right while (j < n - 1 && arr[j] > arr[j + 1]) { y++; j++; } totalPrizes += Math.max( x, y ); } document.write(totalPrizes); return 0; } let arr = [ 1, 2, 2, 3 ]; let n = arr.length; findMinPrizes(arr, n); // This code is contributed by divyesh072019.</script> |
Output:
6
Performance Analysis:
- Time Complexity: O(N2)
- Auxiliary Space: O(1)
Efficient Approach: The idea is to precompute the count of consecutive smaller elements at left and right for every element of the array. It means that, if an element on the left is smaller, then all the smaller elements at the left of that element will also be smaller to the current element. i.e.
if (arr[i-1] < arr[i])
smallerLeft[i] = smallerLeft[i-1] + 1
Similarly, the consecutive smaller elements can be computed using the fact that, if an element on the right is greater than the current element then the consecutive greater elements on the right will also be greater than the current element. i.e.
if (arr[i] < arr[i+1])
smallerRight[i] = smallerRight[i+1] + 1
Below is the implementation of the above approach:
C++
// C++ implementation to find the// minimum prizes required such// that adjacent smaller elements// gets less number of prizes#include <bits/stdc++.h>using namespace std;// Function to find the minimum// number of required such that// adjacent smaller elements gets// less number of prizesint minPrizes(int arr[], int n){ int dpLeft[n]; dpLeft[0] = 1; // Loop to compute the smaller // elements at the left for (int i = 1; i < n; i++) { if (arr[i] > arr[i - 1]) { dpLeft[i] = dpLeft[i - 1] + 1; } else { dpLeft[i] = 1; } } int dpRight[n]; dpRight[n - 1] = 1; // Loop to find the smaller // elements at the right for (int i = n - 2; i >= 0; i--) { if (arr[i] > arr[i + 1]) { dpRight[i] = dpRight[i + 1] + 1; } else { dpRight[i] = 1; } } int totalPrizes = 0; // Loop to find the minimum // prizes required for (int i = 0; i < n; i++) { totalPrizes += max(dpLeft[i], dpRight[i]); } cout << totalPrizes << endl; return 0;}// Driver Codeint main(){ int arr[] = { 1, 2, 2, 3 }; int n = sizeof(arr) / sizeof(arr[0]); minPrizes(arr, n); return 0;} |
Java
// Java implementation to find the// minimum prizes required such// that adjacent smaller elements// gets less number of prizesimport java.util.*;class GFG{// Function to find the minimum// number of required such that// adjacent smaller elements gets// less number of prizesstatic int minPrizes(int arr[], int n){ int []dpLeft = new int[n]; dpLeft[0] = 1; // Loop to compute the smaller // elements at the left for(int i = 1; i < n; i++) { if (arr[i] > arr[i - 1]) { dpLeft[i] = dpLeft[i - 1] + 1; } else { dpLeft[i] = 1; } } int []dpRight = new int[n]; dpRight[n - 1] = 1; // Loop to find the smaller // elements at the right for(int i = n - 2; i >= 0; i--) { if (arr[i] > arr[i + 1]) { dpRight[i] = dpRight[i + 1] + 1; } else { dpRight[i] = 1; } } int totalPrizes = 0; // Loop to find the minimum // prizes required for(int i = 0; i < n; i++) { totalPrizes += Math.max(dpLeft[i], dpRight[i]); } System.out.print(totalPrizes + "\n"); return 0;}// Driver Codepublic static void main(String[] args){ int arr[] = { 1, 2, 2, 3 }; int n = arr.length; minPrizes(arr, n);}}// This code is contributed by Amit Katiyar |
Python3
# Python3 implementation to find the # minimum prizes required such # that adjacent smaller elements # gets less number of prizes# Function to find the minimum # number of required such that # adjacent smaller elements gets # less number of prizes def minPrizes(arr, n): dpLeft = [0] * n dpLeft[0] = 1 # Loop to compute the smaller # elements at the left for i in range(n): if arr[i] > arr[i - 1]: dpLeft[i] = dpLeft[i - 1] + 1 else: dpLeft[i] = 1 dpRight = [0] * n dpRight[-1] = 1 # Loop to find the smaller # elements at the right for i in range(n - 2, -1, -1): if arr[i] > arr[i + 1]: dpRight[i] = dpRight[i + 1] + 1 else: dpRight[i] = 1 totalPrizes = 0 # Loop to find the minimum # prizes required for i in range(n): totalPrizes += max(dpLeft[i], dpRight[i]) print(totalPrizes)# Driver codearr = [ 1, 2, 2, 3 ]n = len(arr) minPrizes(arr, n)# This code is contributed by stutipathak31jan |
C#
// C# implementation to find the// minimum prizes required such// that adjacent smaller elements// gets less number of prizesusing System;class GFG{// Function to find the minimum// number of required such that// adjacent smaller elements gets// less number of prizesstatic int minPrizes(int []arr, int n){ int []dpLeft = new int[n]; dpLeft[0] = 1; // Loop to compute the smaller // elements at the left for(int i = 1; i < n; i++) { if (arr[i] > arr[i - 1]) { dpLeft[i] = dpLeft[i - 1] + 1; } else { dpLeft[i] = 1; } } int []dpRight = new int[n]; dpRight[n - 1] = 1; // Loop to find the smaller // elements at the right for(int i = n - 2; i >= 0; i--) { if (arr[i] > arr[i + 1]) { dpRight[i] = dpRight[i + 1] + 1; } else { dpRight[i] = 1; } } int totalPrizes = 0; // Loop to find the minimum // prizes required for(int i = 0; i < n; i++) { totalPrizes += Math.Max(dpLeft[i], dpRight[i]); } Console.Write(totalPrizes + "\n"); return 0;}// Driver Codepublic static void Main(String[] args){ int []arr = { 1, 2, 2, 3 }; int n = arr.Length; minPrizes(arr, n);}}// This code is contributed by Rajput-Ji |
Javascript
<script>// Javascript implementation to find the// minimum prizes required such// that adjacent smaller elements// gets less number of prizes// Function to find the minimum// number of required such that// adjacent smaller elements gets// less number of prizesfunction minPrizes(arr, n){ let dpLeft = new Array(n); dpLeft[0] = 1; // Loop to compute the smaller // elements at the left for(let i = 1; i < n; i++) { if (arr[i] > arr[i - 1]) { dpLeft[i] = dpLeft[i - 1] + 1; } else { dpLeft[i] = 1; } } let dpRight = new Array(n); dpRight[n - 1] = 1; // Loop to find the smaller // elements at the right for(let i = n - 2; i >= 0; i--) { if (arr[i] > arr[i + 1]) { dpRight[i] = dpRight[i + 1] + 1; } else { dpRight[i] = 1; } } let totalPrizes = 0; // Loop to find the minimum // prizes required for(let i = 0; i < n; i++) { totalPrizes += Math.max(dpLeft[i], dpRight[i]); } document.write(totalPrizes + "</br>"); return 0;}// Driver codelet arr = [ 1, 2, 2, 3 ];let n = arr.length;minPrizes(arr, n);// This code is contributed by suresh07</script> |
Output:
6
Performance Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(N)
Efficient approach : Space optimization O(1)
In previous approach we the current value dpLeft[i] is only depend upon the previous values i.e. dp[i-1] and current value dpRight[i] is only depend upon the next values i.e. dpRight[i+1] So to optimize the space we can keep track of previous , next and current values by the help of variables which will reduce the space complexity from O(N) to O(1).
Implementation Steps:
- Create 2 variables dpLeft and dpRight to keep track of previous and next values of DP.
- Initialize base case dpLeft = dpRight = 1.
- Create a variable totalPrizes to store current value.
- Iterate over subproblem using loop and update totalPrizes.
- After every iteration update dpLeft and dpRight for further iterations.
- At last return totalPrizes.
Implementation
C++
#include <bits/stdc++.h>using namespace std;// Function to find the minimum// number of required such that// adjacent smaller elements gets// less number of prizesint minPrizes(int arr[], int n){ int dpLeft = 1, dpRight = 1, totalPrizes = 1; // Loop to compute the smaller // elements at the left and right for (int i = 1; i < n; i++) { if (arr[i] > arr[i - 1]) { dpLeft++; } else { dpLeft = 1; } if (arr[n - i - 1] > arr[n - i]) { dpRight++; } else { dpRight = 1; } totalPrizes += max(dpLeft, dpRight); } cout << totalPrizes << endl; return 0;}// Driver Codeint main(){ int arr[] = { 1, 2, 2, 3 }; int n = sizeof(arr) / sizeof(arr[0]); minPrizes(arr, n); return 0;} |
Python3
# Python program for above approach# Function to find the minimum# number of required such that# adjacent smaller elements gets# less number of prizesdef minPrizes(arr, n): dpLeft = 1 dpRight = 1 totalPrizes = 1 # Loop to compute the smaller # elements at the left and right for i in range(1, n): if arr[i] > arr[i-1]: dpLeft += 1 else: dpLeft = 1 if arr[n-i-1] > arr[n-i]: dpRight += 1 else: dpRight = 1 totalPrizes += max(dpLeft, dpRight) print(totalPrizes) return 0# Driver Codeif __name__ == '__main__': arr = [1, 2, 2, 3] n = len(arr) minPrizes(arr, n) |
Java
import java.util.*;public class Main { // Function to find the minimum number // of required such that adjacent smaller // elements gets less number of prizes static int minPrizes(int arr[], int n) { int dpLeft = 1, dpRight = 1, totalPrizes = 1; // Loop to compute the smaller // elements at the left and right for (int i = 1; i < n; i++) { if (arr[i] > arr[i - 1]) { dpLeft++; } else { dpLeft = 1; } if (arr[n - i - 1] > arr[n - i]) { dpRight++; } else { dpRight = 1; } totalPrizes += Math.max(dpLeft, dpRight); } System.out.println(totalPrizes); return 0; } // Driver Code public static void main(String[] args) { int arr[] = { 1, 2, 2, 3 }; int n = arr.length; minPrizes(arr, n); }} |
Output
6
Performance Analysis:
- Time Complexity: O(N)
- Auxiliary Space: O(1)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
