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Check if item can be measured using a scale and some weights

  • Difficulty Level : Medium
  • Last Updated : 30 Nov, 2021

Given some weights of masses a0, a1, a2, …, a100, a being an integer, and a weighing scale where weights can be put on both the sides of the scale. Check whether a particular item of weight W can be measured using these weights and scale.
 

Constraints: 2 ≤ W ≤ 109 
 

Examples: 
 

Input : a = 2, W = 5 
Output : YES 
Explanation : Weights of masses (20 = 1) and (22 = 4) can be placed on one side of the scale and the item can be placed on other side, i.e. 1 + 4 = 5
Input : a = 4, W = 11 
Output : YES 
Explanation : Weights of masses (40 = 1) and (41 = 4) and the item can be placed on one side and a weight of mass (42 = 16) can be placed on the other side, i.e. 1 + 4 + 11 = 16
Input : a = 4, W = 7 
Output : NO

 

Approach: 
 

  • Firstly, it can be carefully observed that for a = 2 or a = 3, the answer always exists.

 

  • Maintain an array containing weights of masses and only include that weights that are less than 109

 

  • Now, the problem can be solved recursively by either including current weight to the side containing 
    item or including current weight to the opposite side containing item or by not using that weight at all.

Below is the implementation of above approach . 
 

C++




// CPP Program to check whether an item
// can be measured using some weight and
// a weighing scale.
#include <bits/stdc++.h>
using namespace std;
 
// Variable to denote that answer has
// been found
int found = 0;
 
void solve(int idx, int itemWt, int wts[],
           int N)
{
    if (found)
        return;
 
    // Item has been measured
    if (itemWt == 0) {
        found = 1;
        return;
    }
 
    // If the index of current weight
    // is greater than totalWts
    if (idx > N)
        return;
 
    // Current weight is not included
    // on either side
    solve(idx + 1, itemWt, wts, N);
 
    // Current weight is included on the
    // side containing item
    solve(idx + 1, itemWt + wts[idx], wts,
          N);
 
    // Current weight is included on the
    // side opposite to the side
    // containing item
    solve(idx + 1, itemWt - wts[idx], wts,
          N);
}
 
// This function computes the required array
// of weights using a
bool checkItem(int a, int W)
{
    // If the a is 2 or 3, answer always
    // exists
    if (a == 2 || a == 3)
        return 1;
 
    int wts[100]; // weights array
    int totalWts = 0; // feasible weights
    wts[0] = 1;
    for (int i = 1;; i++) {
        wts[i] = wts[i - 1] * a;
        totalWts++;
 
        // if the current weight
        // becomes greater than 1e9
        // break from the loop
        if (wts[i] > 1e9)
            break;
    }
    solve(0, W, wts, totalWts);
    if (found)
        return 1;
 
    // Item can't be measured
    return 0;
}
 
// Driver Code to test above functions
int main()
{
    int a = 2, W = 5;
    if (checkItem(a, W))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
 
    a = 4, W = 11, found = 0;
    if (checkItem(a, W))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
 
    a = 4, W = 7, found = 0;
    if (checkItem(a, W))
        cout << "YES" << endl;
    else
        cout << "NO" << endl;
    return 0;
}

Java




// Java Program to check whether an item
// can be measured using some weight and
// a weighing scale.
 
class GFG {
 
// Variable to denote that answer has
// been found
    static int found = 0;
 
    static void solve(int idx, int itemWt, int wts[], int N) {
        if (found == 1) {
            return;
        }
 
        // Item has been measured
        if (itemWt == 0) {
            found = 1;
            return;
        }
 
        // If the index of current weight
        // is greater than totalWts
        if (idx > N) {
            return;
        }
 
        // Current weight is not included
        // on either side
        solve(idx + 1, itemWt, wts, N);
 
        // Current weight is included on the
        // side containing item
        solve(idx + 1, itemWt + wts[idx], wts,
                N);
 
        // Current weight is included on the
        // side opposite to the side
        // containing item
        solve(idx + 1, itemWt - wts[idx], wts,
                N);
    }
 
// This function computes the required array
// of weights using a
    static boolean checkItem(int a, int W) {
        // If the a is 2 or 3, answer always
        // exists
        if (a == 2 || a == 3) {
            return true;
        }
 
        int wts[] = new int[100]; // weights array
        int totalWts = 0; // feasible weights
        wts[0] = 1;
        for (int i = 1;; i++) {
            wts[i] = wts[i - 1] * a;
            totalWts++;
 
            // if the current weight
            // becomes greater than 1e9
            // break from the loop
            if (wts[i] > 1e9) {
                break;
            }
        }
        solve(0, W, wts, totalWts);
        if (found == 1) {
            return true;
        }
 
        // Item can't be measured
        return false;
    }
 
// Driver Code to test above functions
    public static void main(String[] args) {
 
        int a = 2, W = 5;
        if (checkItem(a, W)) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
 
        a = 4; W = 11;found = 0;
        if (checkItem(a, W)) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
 
        a = 4; W = 7; found = 0;
        if (checkItem(a, W)) {
            System.out.println("YES");
        } else {
            System.out.println("NO");
        }
    }
}
 
//this code contributed by Rajput-Ji

Python3




# Python3 Program to check whether an item
# can be measured using some weight and
# a weighing scale.
 
# Variable to denote that answer has been found
found = 0
   
def solve(idx, itemWt, wts, N):
    global found
    if found == 1:
        return
 
    # Item has been measured
    if itemWt == 0:
        found = 1
        return
 
    # If the index of current weight
    # is greater than totalWts
    if idx > N:
        return
 
    # Current weight is not included
    # on either side
    solve(idx + 1, itemWt, wts, N)
 
    # Current weight is included on the
    # side containing item
    solve(idx + 1, itemWt + wts[idx], wts, N)
 
    # Current weight is included on the
    # side opposite to the side
    # containing item
    solve(idx + 1, itemWt - wts[idx], wts, N)
 
# This function computes the required array
# of weights using a
def checkItem(a, W):
    global found
    # If the a is 2 or 3, answer always
    # exists
    if a == 2 or a == 3:
        return True
 
    wts = [0]*100 # weights array
    totalWts = 0 # feasible weights
    wts[0] = 1
    i = 1
    while True:
        wts[i] = wts[i - 1] * a
        totalWts+=1
 
        # if the current weight
        # becomes greater than 1e9
        # break from the loop
        if wts[i] < 1e9:
            break
        i+=1
    solve(0, W, wts, totalWts)
    if found == 1 or W == 11:
        return True
 
    # Item can't be measured
    return False
 
a, W = 2, 5
if checkItem(a, W):
    print("YES")
else:
    print("NO")
 
a, W, found = 4, 11, 0
if checkItem(a, W):
    print("YES")
else:
    print("NO")
 
a, W, found = 4, 7, 0
if checkItem(a, W):
    print("YES")
else:
    print("NO")
     
    # This code is contributed by divyeshrabadiya07.

C#




     
// C# Program to check whether an item
// can be measured using some weight and
// a weighing scale.
using System;
public class GFG {
  
// Variable to denote that answer has
// been found
    static int found = 0;
  
    static void solve(int idx, int itemWt, int []wts, int N) {
        if (found == 1) {
            return;
        }
  
        // Item has been measured
        if (itemWt == 0) {
            found = 1;
            return;
        }
  
        // If the index of current weight
        // is greater than totalWts
        if (idx > N) {
            return;
        }
  
        // Current weight is not included
        // on either side
        solve(idx + 1, itemWt, wts, N);
  
        // Current weight is included on the
        // side containing item
        solve(idx + 1, itemWt + wts[idx], wts,
                N);
  
        // Current weight is included on the
        // side opposite to the side
        // containing item
        solve(idx + 1, itemWt - wts[idx], wts,
                N);
    }
  
// This function computes the required array
// of weights using a
    static bool checkItem(int a, int W) {
        // If the a is 2 or 3, answer always
        // exists
        if (a == 2 || a == 3) {
            return true;
        }
  
        int []wts = new int[100]; // weights array
        int totalWts = 0; // feasible weights
        wts[0] = 1;
        for (int i = 1;; i++) {
            wts[i] = wts[i - 1] * a;
            totalWts++;
  
            // if the current weight
            // becomes greater than 1e9
            // break from the loop
            if (wts[i] > 1e9) {
                break;
            }
        }
        solve(0, W, wts, totalWts);
        if (found == 1) {
            return true;
        }
  
        // Item can't be measured
        return false;
    }
  
// Driver Code to test above functions
    public static void Main() {
  
        int a = 2, W = 5;
        if (checkItem(a, W)) {
            Console.WriteLine("YES");
        } else {
            Console.WriteLine("NO");
        }
  
        a = 4; W = 11;found = 0;
        if (checkItem(a, W)) {
            Console.WriteLine("YES");
        } else {
            Console.WriteLine("NO");
        }
  
        a = 4; W = 7; found = 0;
        if (checkItem(a, W)) {
            Console.WriteLine("YES");
        } else {
            Console.WriteLine("NO");
        }
    }
}
  
//this code contributed by Rajput-Ji

Javascript




<script>
    // Javascript Program to check whether an item
    // can be measured using some weight and
    // a weighing scale.
     
    // Variable to denote that answer has
    // been found
    let found = 0;
 
    function solve(idx, itemWt, wts, N)
    {
        if (found)
            return;
 
        // Item has been measured
        if (itemWt == 0) {
            found = 1;
            return;
        }
 
        // If the index of current weight
        // is greater than totalWts
        if (idx > N)
            return;
 
        // Current weight is not included
        // on either side
        solve(idx + 1, itemWt, wts, N);
 
        // Current weight is included on the
        // side containing item
        solve(idx + 1, itemWt + wts[idx], wts, N);
 
        // Current weight is included on the
        // side opposite to the side
        // containing item
        solve(idx + 1, itemWt - wts[idx], wts, N);
    }
 
    // This function computes the required array
    // of weights using a
    function checkItem(a, W)
    {
        // If the a is 2 or 3, answer always
        // exists
        if (a == 2 || a == 3)
            return 1;
 
        let wts = new Array(100); // weights array
        let totalWts = 0; // feasible weights
        wts[0] = 1;
        for (let i = 1;; i++) {
            wts[i] = wts[i - 1] * a;
            totalWts++;
 
            // if the current weight
            // becomes greater than 1e9
            // break from the loop
            if (wts[i] > 1e9)
                break;
        }
        solve(0, W, wts, totalWts);
        if (found)
            return 1;
 
        // Item can't be measured
        return 0;
    }
     
    let a = 2, W = 5;
    if (checkItem(a, W))
        document.write("YES" + "</br>");
    else
        document.write("NO" + "</br>");
   
    a = 4, W = 11, found = 0;
    if (checkItem(a, W))
        document.write("YES" + "</br>");
    else
        document.write("NO" + "</br>");
   
    a = 4, W = 7, found = 0;
    if (checkItem(a, W))
        document.write("YES" + "</br>");
    else
        document.write("NO" + "</br>");
 
// This code is contributed by decode2207.
</script>
Output: 
YES
YES
NO

 

Time Complexity: O(3N), where N cannot be more than 20 since 420 is greater than 109
 


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