Given N objects numbered from 1 to N out of which all are of the same weights except only one object which is not known beforehand. We are also given Q comparisons, in each of which an equal number of objects are placed on both sides of a balance scale, and we are told the heavier side.
The task is to find the inconsistently weighted object or determine if the data is not sufficient enough.
Examples:
Input : N = 6 Q = 3 1 2 = 5 6 1 2 3 > 4 5 6 3 4 < 5 6 Output : 4 Explanation: Object 4 is lighter than all other objects. Input : N = 10 Q = 4 1 2 3 4 < 7 8 9 10 1 = 9 2 3 4 > 1 5 10 6 = 2 Output : Insufficient data
It is told that except only one element, the rest of the elements are of the same weights. So, if we observe carefully, it can be said that:
- In a ‘=’ comparison, none of the objects on both sides is the inconsistently weighted one.
- If an object appears on the heavier side in one comparison and on the lighter side in another, then it is not the inconsistently weighted object. This is because, if an object appears on the heavier side then it is of the maximum weight and if it appears on the lighter side then it is of the minimum weight. Since a single element can’t be both maximum and minimum at the same time. So, this case will never occur.
- The inconsistently weighted object must appear in all of the non-balanced (‘>’ or ‘<‘) comparisons.
We can use the above three observations to narrow down the potential candidates for the inconsistently weighted object. We will consider only those objects which are either on the heavier side or the lighter side; if there is only one such object then it is the required one. If there is no such object, then we will consider all those objects which do not appear in any comparison. If there is only one such object then it is the inconsistently weighted object. If none of these scenarios arises, the data is insufficient.
Below is the implementation of the above approach:
Python3
# Python program to determine the # inconsistently weighted object # Function to get the difference of two lists def subt(A, B): return list(set(A) - set(B)) # Function to get the intersection of two lists def intersection(A, B): return list(set(A).intersection(set(B))) # Function to get the intersection of two lists def union(A, B): return list(set(A).union(set(B))) # Function to find the inconsistently weighted object def inconsistentlyWeightedObject(N, Q, comparisons): # Objects which appear on the heavier side heavierObj = [i for i in range(1, N + 1)] # Objects which appear on the lighter side lighterObj = [i for i in range(1, N + 1)] equalObj = [] # Objects which appear in '=' comparisons # Objects which don't appear in any comparison objectNotCompared = [i for i in range(1, N + 1)] for c in comparisons: objectNotCompared = subt(objectNotCompared, union(c[0], c[2])) if c[1] == '=': equalObj = union(equalObj, union(c[0], c[2])) elif c[1] == '<': # Removing those objects which do # not appear on the lighter side lighterObj = intersection(lighterObj, c[0]) # Removing those objects which do # not appear on the heavier side heavierObj = intersection(heavierObj, c[2]) else: # Removing those objects which do # not appear on the lighter side lighterObj = intersection(lighterObj, c[2]) # Removing those objects which do # not appear on the heavier side heavierObj = intersection(heavierObj, c[0]) L_iwo = subt(union(heavierObj, lighterObj), equalObj) # Potential candidates if len(L_iwo) == 1: return L_iwo[0] elif not len(L_iwo): if len(objectNotCompared) == 1: return objectNotCompared[0] else: return 'Insufficient data' else: return 'Insufficient data'# Driver code N = 6Q = 3comparisons = [ [[1, 2], '=', [5, 6]], [[1, 2, 3], '>', [4, 5, 6]], [[3, 4], '<', [5, 6]] ] print(inconsistentlyWeightedObject(N, Q, comparisons)) |
Javascript
// JavaScript program to determine the // inconsistently weighted object // Function to get the difference of two lists function subt(A, B){ return A.filter(x => !B.includes(x))}// Function to get the intersection of two lists function intersection(A, B){ return A.filter(x => B.includes(x));}// Function to get the intersection of two lists function union(A, B){ return [...A, ...B]; }// Function to find the inconsistently weighted object function inconsistentlyWeightedObject(N, Q, comparisons){ // Objects which appear on the heavier side let heavierObj = []; for (i = 1; i <= N; i++) heavierObj.push(i); // Objects which appear on the lighter side let lighterObj = []; for (i = 1; i <= N; i++) lighterObj.push(i); let equalObj = []; // Objects which appear in '=' comparisons // Objects which don't appear in any comparison objectNotCompared = []; for (i = 1; i <= N; i++) objectNotCompared.push(i); for (var c of comparisons) { objectNotCompared = subt(objectNotCompared, union(c[0], c[2])) ; if (c[1] == '=') equalObj = union(equalObj, union(c[0], c[2])) else if (c[1] == '<') { // Removing those objects which do // not appear on the lighter side lighterObj = intersection(lighterObj, c[0]) // Removing those objects which do // not appear on the heavier side heavierObj = intersection(heavierObj, c[2]) } else { // Removing those objects which do // not appear on the lighter side lighterObj = intersection(lighterObj, c[2]) // Removing those objects which do // not appear on the heavier side heavierObj = intersection(heavierObj, c[0]) } } let L_iwo = subt(union(heavierObj, lighterObj), equalObj) // Potential candidates if ((L_iwo).length == 1) return L_iwo[0] else if ((L_iwo).length == 0) { if (objectNotCompared.length == 1) return objectNotCompared[0] else return 'Insufficient data' } else return 'Insufficient data'}// Driver code let N = 6let Q = 3let comparisons = [ [[1, 2], '=', [5, 6]], [[1, 2, 3], '>', [4, 5, 6]], [[3, 4], '<', [5, 6]] ] console.log(inconsistentlyWeightedObject(N, Q, comparisons)) // This code is contributed by phasing17 |
Output:
4
Time Complexity: O(nlogn)
Auxiliary Space: O(n)
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