Chocolate Distribution Problem | Set 2 - GeeksforGeeks

14 Dec, 2021
Given an array A[] consisting of N integers, where each value represents the marks of the ith student, the task is to find the minimum number of chocolates required to be distributed such that:
Each student should be awarded with at least one chocolate
A student with higher marks should be awarded more chocolates than his adjacent students.
Examples:
Input: A[] = {10, 30, 20}Output: 4Explanation : Since, 30 is larger than its adjacent, so the second student must get more chocolates. Therefore, the minimum chocolates can be distributed as {1, 2, 1} = 1 + 2 + 1 = 4
Input: A[] = {23, 14, 15, 14, 56, 29, 14}Output: 12 
Method 1:
Approach: The problem can be solved using Greedy approach. Follow the steps below to solve the problem:
Initialize array B[] of length N with 1.
Traverse from left to right from i = 1 to N – 1, updating B[i] as B[i] = B[i-1]+1 if A[i] greater the A[i-1].
After completing the above step, traverse again from right to left from i = N – 2 to 0, updating B[i] as B[i] = max(B[i], B[i+1]+1) if A[i] is greater than A[i + 1]. Otherwise, update B[i] as B[i] = max(B[i], 1).
After traversing, calculate the sum of the array B[] and print it as the minimum number of candies required.
Below is the implementation of the above approach:
C++
Java
Python3
C#
Javascript
// C++ program for the above approach  #include <iostream>using namespace std;  // FUnction to print minimum number// of candies requiredvoid minChocolates(int A[], int N){    int B[N];      // Distribute 1 chocolate to each    for (int i = 0; i < N; i++) {        B[i] = 1;    }      // Traverse from left to right    for (int i = 1; i < N; i++) {        if (A[i] > A[i - 1])            B[i] = B[i - 1] + 1;        else            B[i] = 1;    }      // Traverse from right to left    for (int i = N - 2; i >= 0; i--) {        if (A[i] > A[i + 1])            B[i] = max(B[i + 1] + 1, B[i]);        else            B[i] = max(B[i], 1);    }      // Initialize sum    int sum = 0;      // Find total sum    for (int i = 0; i < N; i++) {        sum += B[i];    }      // Return sum    cout << sum << "\n";}  // Driver Codeint main(){      // Given array    int A[] = { 23, 14, 15, 14, 56, 29, 14 };      // Size of the given array    int N = sizeof(A) / sizeof(A[0]);      minChocolates(A, N);}
// Java program for the above approachimport java.util.*;class GFG {      // FUnction to print minimum number    // of candies required    static void minChocolates(int A[], int N)    {        int[] B = new int[N];          // Distribute 1 chocolate to each        for (int i = 0; i < N; i++) {            B[i] = 1;        }          // Traverse from left to right        for (int i = 1; i < N; i++) {            if (A[i] > A[i - 1])                B[i] = B[i - 1] + 1;            else                B[i] = 1;        }          // Traverse from right to left        for (int i = N - 2; i >= 0; i--) {            if (A[i] > A[i + 1])                B[i] = Math.max(B[i + 1] + 1, B[i]);            else                B[i] = Math.max(B[i], 1);        }          // Initialize sum        int sum = 0;          // Find total sum        for (int i = 0; i < N; i++) {            sum += B[i];        }          // Return sum        System.out.print(sum + "\n");    }      // Driver Code    public static void main(String[] args)    {          // Given array        int A[] = { 23, 14, 15, 14, 56, 29, 14 };          // Size of the given array        int N = A.length;        minChocolates(A, N);    }}  // This code contributed by shikhasingrajput
# Python3 program for the above approach  # Function to print minimum number# of candies required    def minChocolates(A, N):      B = [1 for i in range(N)]      # Traverse from left to right    for i in range(1, N):        if (A[i] > A[i - 1]):            B[i] = B[i - 1] + 1        else:            B[i] = 1      # Traverse from right to left    for i in range(N - 2, -1, -1):        if (A[i] > A[i + 1]):            B[i] = max(B[i + 1] + 1, B[i])        else:            B[i] = max(B[i], 1)      # Initialize sum    sum = 0      # Find total sum    for i in range(N):        sum += B[i]      # Return sum    print(sum)    # Driver Codeif __name__ == '__main__':      # Given array    A = [23, 14, 15, 14,         56, 29, 14]      # Size of the given array    N = len(A)      minChocolates(A, N)  # This code is contributed by mohit kumar 29
// C# program for the above approachusing System;public class GFG {      // FUnction to print minimum number    // of candies required    static void minChocolates(int[] A, int N)    {        int[] B = new int[N];          // Distribute 1 chocolate to each        for (int i = 0; i < N; i++) {            B[i] = 1;        }          // Traverse from left to right        for (int i = 1; i < N; i++) {            if (A[i] > A[i - 1])                B[i] = B[i - 1] + 1;            else                B[i] = 1;        }          // Traverse from right to left        for (int i = N - 2; i >= 0; i--) {            if (A[i] > A[i + 1])                B[i] = Math.Max(B[i + 1] + 1, B[i]);            else                B[i] = Math.Max(B[i], 1);        }          // Initialize sum        int sum = 0;          // Find total sum        for (int i = 0; i < N; i++) {            sum += B[i];        }          // Return sum        Console.Write(sum + "\n");    }      // Driver Code    public static void Main(String[] args)    {          // Given array        int[] A = { 23, 14, 15, 14, 56, 29, 14 };          // Size of the given array        int N = A.Length;        minChocolates(A, N);    }}  // This code is contributed by 29AjayKumar
<script>  // javascript program for the above approach      // FUnction to print minimum number    // of candies required    function minChocolates(A, N)    {        let B = new Array(N).fill(0);          // Distribute 1 chocolate to each        for (let i = 0; i < N; i++) {            B[i] = 1;        }          // Traverse from left to right        for (let i = 1; i < N; i++) {            if (A[i] > A[i - 1])                B[i] = B[i - 1] + 1;            else                B[i] = 1;        }          // Traverse from right to left        for (let i = N - 2; i >= 0; i--) {            if (A[i] > A[i + 1])                B[i] = Math.max(B[i + 1] + 1, B[i]);            else                B[i] = Math.max(B[i], 1);        }          // Initialize sum        let sum = 0;          // Find total sum        for (let i = 0; i < N; i++) {            sum += B[i];        }          // Return sum        document.write(sum + "<br/>");    }  // Driver Code          // Given array        let A = [ 23, 14, 15, 14, 56, 29, 14 ];          // Size of the given array        let N = A.length;        minChocolates(A, N);  </script>
12
 
Time Complexity: O(N) where N is the length of the given array.Auxiliary Space: O(N)
Method 2 : Efficient approach
On careful observation, the space complexity can be reduced to O(1).
I. Observation:
Marks array will be a combination of strictly increasing, strictly decreasing or flat (value is same as both the neighbors) subarrays.
To minimize the total number of chocolates distributed, the number of chocolates received by a person and at least one of the neighbors should **differ by 1 or less.
** An exception of second observation is mentioned below
II. Distributing chocolates
Case 1: subarray is strictly increasing
If the values are strictly increasing, number of chocolates given to ith student will be one more than the number of chocolates given to (i-1)th  student (for any i > 0)
One chocolate will be given to the left most person in subarray, two the next and so on incrementally up to the person with highest marks.
For a strictly increasing subarray of length k, the chocolate distribution will be [1, 2, ... , k].
Case 2 : subarray is strictly decreasing
Number of chocolates given to ith student will be one more than the chocolates given to (i+1)th  student ( for any i < n-1) with one chocolate to the rightmost person and max number to the leftmost person of the subarray.
For a strictly decreasing subarray of length k, the chocolate distribution will be [k, k-1, ... ,1].
Case 3 : flat sequence
Given that a student with highest marks will be awarded more number of chocolates than neighbors. So there is no dependency if the values are equal. Minimum value will be assigned for optimal result.
One chocolate will be given to person at position i if both the adjacent values are equal to a[i] i.e, a[i-1] == a[i] == a[i+1]
For a flat subarray of length k, the chocolate distribution will be [1, 1, ... ,1].
**The difference for an assigned value with both the neighbors may be greater than 1, if there is a single element in flat sequence and it lies exactly between an increasing & decreasing sequence
Transition points: The points where the trend(increasing/ decreasing/ flat nature) of subarray changes.
Peak point : Point which is end point of one increasing sequence and start point of other decreasing sequence
then value assigned will be max(k1, k2)
where k1 – value obtained from increasing sequence,
k2 – value obtained from decreasing sequence.
This point will be considered as part of either increasing or decreasing sequence only 
III. Result : 
 As the values in an increasing/decreasing sequence differ by 1, the number of chocolates distributed to students in a specific subarray of k elements will be  sum of k natural numbers. And the count will be k for a flat sequence as all the values are 1. The required value will be the total sum of the results of subarrays.
IV. Implementation:
Consider variables i, j initially pointing to first element, val = 1, res = 0.
After traversing through the array res gives the total number of chocolates distributed.
val while iterating index j (in increasing/flat subarray) represents the number of chocolates received by the person at j
If the subarray is increasing or a flat sequence, val is added to res; i, j are moved forward and val is updated according to next value (a[j + 1]).
If the subarray is decreasing, i is pointed to the starting point of subarray and j is moved forward till next transition point. val, res are not updated till the end of the subarray. In this case val holds the value of the peak element obtained from previous subarray. At the end of the decreasing sequence res is updated using get_sum function & val is updated to point to the number of chocolates held by the next person.
V. Example:
Input: A[ ] = {1, 2, 10, 7, 6, 4, 5, 5, 5, 6}
Output : 19
Explanation:
subarray — sequence type    —  count of chocolates
A[0-1] — increasing sequence — [1, 2]
A[2-5] — decreasing sequence — [4, 3, 2, 1]
A[5-6] — increasing sequence — [1, 2]
A[7-7] — flat sequence       — [1]
A[8-9] — increasing sequence — [1, 2]
A[2], A[9] are peak points
Chocolates distribution will be 
[1, 2, 4, 3, 2, 1, 2, 1, 1, 2]
Sum of all values = 19
Below is the code for above approach
C
C++
Java
Python3
Javascript
C#
// C program for above approach#include <stdio.h>  // Helper function to get sum of decreasing sequenceint get_sum(int peak, int start, int end){    /* peak is the value obtained at peak point       from previous flat/increasing sequence */      /* value obtained from decreasing sequence     also the count of values in the sequence*/    int count = end - start + 1;      /* assigning max of values obtained from     increasing and decreasing sequences */    peak = (peak > count) ? peak : count;      /* sum of count - 1 values & peak value     sum of natural numbers : (n * (n + 1))/2 */    int s = peak + (((count - 1) * count) >> 1);      return s;}  // Function to return minimum number of chocolatesint minChocolates(int a[], int n){    int i = 0, j = 0;    int res = 0, val = 1;      while (j < n - 1) {          if (a[j] > a[j + 1]) {            // decreasing sequence            j += 1;            continue;        }          if (i == j)            // add the chocolates received by that person            res += val;        else {            // end point of decreasing sequence            res += get_sum(val, i, j);            val = 1; // reset value at that index        }          if (a[j] < a[j + 1])            // increasing sequence            val += 1;        else            // flat sequence            val = 1;          j += 1;        i = j;    }    // add value of chocolates at position n-1    if (i == j)        res += val;    else        res += get_sum(val, i, j);      return res;}  // Driver codeint main(){      int a[] = { 5, 5, 4, 3, 2, 1 };    int n = sizeof(a) / sizeof(a[0]);    printf("Minimum number of chocolates = %d",           minChocolates(a, n));    return 0;}  // This code is contributed by saitejagampala
// C++ program for above approach  #include <iostream>using namespace std;  // Helper function to get sum of decreasing sequenceint get_sum(int peak, int start, int end){    /* peak is the value obtained at peak point       from previous flat/increasing sequence */      /* value obtained from decreasing sequence     also the count of values in the sequence*/    int count = end - start + 1;      /* assigning max of values obtained from     increasing and decreasing sequences */    peak = max(peak, count);      /* sum of count - 1 values & peak value     sum of natural numbers : (n * (n + 1))/2 */    int s = peak + (((count - 1) * count) >> 1);      return s;}  // Function to return minimum number of chocolatesint minChocolates(int a[], int n){    int i = 0, j = 0;    int res = 0, val = 1;      while (j < n - 1) {          if (a[j] > a[j + 1]) {            // decreasing sequence            j += 1;            continue;        }          if (i == j)            // add the chocolates received by that person            res += val;        else {            // end point of decreasing sequence            res += get_sum(val, i, j);            val = 1; // reset value at that index        }          if (a[j] < a[j + 1])            // increasing sequence            val += 1;        else            // flat sequence            val = 1;          j += 1;        i = j;    }    // add value of chocolates at position n-1    if (i == j)        res += val;    else        res += get_sum(val, i, j);      return res;}  // Driver codeint main(){      int a[] = { 5, 5, 4, 3, 2, 1 };    int n = sizeof(a) / sizeof(a[0]);    cout << "Minimum number of chocolates = "         << minChocolates(a, n) << "\n";    return 0;}  // This code is contributed by saitejagampala
// Java program for above approach  import java.io.*;  class GFG {    public static void main(String[] args)    {        int[] a = { 5, 5, 4, 3, 2, 1 };        int n = a.length;        System.out.print("Minimum number of chocolates = "                         + minChocolates(a, n));    }      // Function to return minimum number of chocolates    public static int minChocolates(int[] a, int n)    {        int i = 0, j = 0;        int res = 0, val = 1;          while (j < n - 1) {              if (a[j] > a[j + 1]) {                // decreasing sequence                j += 1;                continue;            }              if (i == j)                // add the chocolates received by that                // person                res += val;            else {                // end point of decreasing sequence                res += get_sum(val, i, j);                val = 1; // reset value at that index            }              if (a[j] < a[j + 1])                // increasing sequence                val += 1;            else                // flat sequence                val = 1;              j += 1;            i = j;        }          // add value of chocolates at position n-1        if (i == j)            res += val;        else            res += get_sum(val, i, j);          return res;    }      // helper function to get sum of decreasing sequence    public static int get_sum(int peak, int start, int end)    {        /* peak is the value obtained at peak point           from previous flat/increasing sequence */          /* value obtained from decreasing sequence         also the count of values in the sequence*/        int count = end - start + 1;          /* assigning max of values obtained from         increasing and decreasing sequences */        peak = (peak > count) ? peak : count;          /* sum of count - 1 values & peak value         sum of natural numbers : (n * (n + 1))/2 */        int s = peak + (((count - 1) * count) >> 1);          return s;    }}  // This code is contributed by saitejagampala
# Python3 program for above approach  # Function to return minimum number of chocolates    def minChocolates(a, n):    i, j = 0, 0    val, res = 1, 0      while(j < n - 1):          if(a[j] > a[j + 1]):            # decreasing sequence            j += 1            continue          if(i == j):            # add the chocolates received by that person            res += val        else:            # end point of decreasing sequence            res += get_sum(val, i, j)            val = 1  # reset value at that index          if(a[j] < a[j + 1]):            # increasing sequence            val += 1        else:            # flat sequence            val = 1          j += 1        i = j      # add value of chocolates at position n-1    if(i == j):        res += val    else:        res += get_sum(val, i, j)      return res    # Helper function to get sum of decreasing sequencedef get_sum(peak, start, end):    # peak is the value obtained at peak point    # from previous flat/increasing sequence      # value obtained from decreasing sequence    # also the count of values in the sequence    count = end - start + 1      # assigning max of values obtained from increasing    # and decreasing sequences    peak = max(peak, count)      # sum of count - 1 values & peak value    # sum of natural numbers : (n * (n + 1))/2    s = peak + (((count-1) * count) >> 1)      return s    # Driver codeif __name__ == '__main__':    a = [5, 5, 4, 3, 2, 1]    n = len(a)    print('Minimum number of chocolates =', minChocolates(a, n))   # This code is contributed by saitejagampala
<script>// Javascript program for above approach  // Function to return minimum number of chocolatesfunction minChocolates(a,n){    let i = 0, j = 0;        let res = 0, val = 1;           while (j < n - 1)         {             if (a[j] > a[j + 1])            {                              // decreasing sequence                j += 1;                continue;            }               if (i == j)                              // add the chocolates received by that                // person                res += val;            else             {                              // end point of decreasing sequence                res += get_sum(val, i, j);                val = 1; // reset value at that index            }               if (a[j] < a[j + 1])                // increasing sequence                val += 1;            else                // flat sequence                val = 1;               j += 1;            i = j;        }           // add value of chocolates at position n-1        if (i == j)            res += val;        else            res += get_sum(val, i, j);           return res;}  // helper function to get sum of decreasing sequencefunction get_sum(peak,start,end){    /* peak is the value obtained at peak point           from previous flat/increasing sequence */           /* value obtained from decreasing sequence         also the count of values in the sequence*/        let count = end - start + 1;           /* assigning max of values obtained from         increasing and decreasing sequences */        peak = (peak > count) ? peak : count;           /* sum of count - 1 values & peak value         sum of natural numbers : (n * (n + 1))/2 */        let s = peak + (((count - 1) * count) >> 1);           return s;}  let a = [5, 5, 4, 3, 2, 1];let n = a.length;document.write("Minimum number of chocolates = "                         + minChocolates(a, n));  // This code is contributed by unknown2108</script>
// C# program for above approach  using System;  public class GFG{          // Function to return minimum number of chocolates    public static int minChocolates(int[] a, int n)    {        int i = 0, j = 0;        int res = 0, val = 1;           while (j < n - 1) {               if (a[j] > a[j + 1]) {                // decreasing sequence                j += 1;                continue;            }               if (i == j)                // add the chocolates received by that                // person                res += val;            else {                // end point of decreasing sequence                res += get_sum(val, i, j);                val = 1; // reset value at that index            }               if (a[j] < a[j + 1])                // increasing sequence                val += 1;            else                // flat sequence                val = 1;               j += 1;            i = j;        }           // add value of chocolates at position n-1        if (i == j)            res += val;        else            res += get_sum(val, i, j);           return res;    }       // helper function to get sum of decreasing sequence    public static int get_sum(int peak, int start, int end)    {        /* peak is the value obtained at peak point           from previous flat/increasing sequence */           /* value obtained from decreasing sequence         also the count of values in the sequence*/        int count = end - start + 1;           /* assigning max of values obtained from         increasing and decreasing sequences */        peak = (peak > count) ? peak : count;           /* sum of count - 1 values & peak value         sum of natural numbers : (n * (n + 1))/2 */        int s = peak + (((count - 1) * count) >> 1);           return s;    }          static public void Main (){                  int[] a = { 5, 5, 4, 3, 2, 1 };        int n = a.Length;        Console.WriteLine("Minimum number of chocolates = "                         + minChocolates(a, n));              }}  // This code is contributed by patel2127
Minimum number of chocolates = 16
Time Complexity : O(N),  N is the length of the array
Space Complexity : O(1)
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