add problem Coin Piles

Author: robotenique

coin-piles/problem_statement.txt

@@ -0,0 +1,46 @@
+Coin Piles
+=======================
+There are N piles of coins each containing  Ai (1<=i<=N) coins.
+Now, you have to adjust the number of coins in each pile such that for any two pile, if a is the number of coins in first pile and b is the number of coins in second pile then |a-b|<= K.
+In order to do that you can remove coins from different piles to decrease the number of coins in those piles but you cannot increase the number of coins in a pile by adding more coins. Now, given a value of N and K, along with the sizes of the N different piles you have to tell the minimum number of coins to be removed in order to satisfy the given condition.
+
+Note: You can also remove a pile by removing all the coins of that pile.
+
+Input
+=====
+
+The first line of the input contains T, the number of test cases. Then T lines follow. Each test case contains two lines. The first line of a test case contains N and K. The second line of the test case contains N integers describing the number of coins in the N piles.
+
+Output
+=====
+
+For each test case output a single integer containing the minimum number of coins needed to be removed in a new line.
+
+Constraints:
+1<=T<=50
+1<=N<=100
+1<=Ai<=1000
+0<=K<=1000
+
+Example
+=====
+
+Input
+3
+4 0
+2 2 2 2
+6 3
+1 2 5 1 1 1
+6 3
+1 5 1 2 5 1
+
+Output
+0
+1
+2
+
+Explanation:
+
+1. In the first test case, for any two piles the difference in the number of coins is <=0. So no need to remove any coins.
+2. In the second test case if we remove one coin from pile containing 5 coins then for any two piles the absolute difference in the number of coins is <=3.
+3. In the third test case if we remove one coin each from both the piles containing 5 coins , then for any two piles the absolute difference in the number of coins is <=3.

coin-piles/solution.cpp

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+#include <bits/stdc++.h>
+using namespace std;
+
+int main() {
+	int t,n,k,a[100],d[100],rem[100],low,hi,removals,d_idx,min_removals;
+	unordered_map<int,int>freq;
+
+	scanf("%d\n", &t);
+	while(t--) {
+	    scanf("%d %d\n", &n, &k);
+	    for(int lc = 0; lc < n; lc++) {
+	        scanf("%d" , &a[lc]);
+	    }
+
+	    sort(a,a + n);
+	    d_idx=0;
+        freq.clear();
+	    for(int lc = 0; lc < n; lc++) {
+	        int key = a[lc], cnt = 0;
+	        while(lc < n && a[lc] == key)
+	            lc++;
+	        lc--;
+	        d[d_idx] = key;
+            d_idx++;
+	    }
+
+	    low = 0;
+        min_removals = INT_MAX;
+	    while(low < d_idx) {
+	        int cur = d[low];
+            removals = 0;
+	        for(int lc = 0; lc < n; lc++) {
+	            if(a[lc] < cur)
+	                rem[lc] = a[lc];
+	            else if(a[lc] >= cur && a[lc] <= (cur + k))
+	                rem[lc] = 0;
+	            else if(a[lc] > (cur + k))
+	                rem[lc] = (a[lc] - (cur + k));
+
+	            removals += rem[lc];
+	        }
+	        min_removals = min(min_removals, removals);
+	        low++;
+	    }
+	    printf("%d\n",min_removals);
+	}
+	return 0;
+}