CareerCup

Typical Dynamic Programming

public void printSums(int c1, int c2, int c3) {

        Set<Integer> sums = new HashSet<>();
        sums.add(0);

        for(int sum = 1; sum <= 1000; sum++) {

            if(sums.contains(sum - c1) || sums.contains(sum - c2) || sums.contains(sum - c3)) {
                System.out.println(sum);
                sums.add(sum);
            }
        }
    }

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Alternative approach:

def count_coins(cs, limit=1000):
    ans = [0]
    for c in cs:
        if c > limit:
            continue
        for x in ans:
            if x+c > limit:
                break
            ans += [x+c]
    return sorted(set(ans))[1:]

void printCoinsUpTo(const vector<int> &coins, int val) {
    vector<int> result;
    vector<int> idx(coins.size(), 0);
    result.push_back(0);

    for (int i = 1; i <= val; ++i) {
        int minVal = INT_MAX;

        for (int i = 0; i < coins.size(); ++i) {
            minVal = min(minVal, coins[i] + result[idx[i]]);
        }
        result.push_back(minVal);

        for (int i = 0; i < coins.size(); ++i) {
            if (coins[i] + result[idx[i]] == minVal) {
                idx[i]++;
            }
        }
    }

    for(int i = 1; i < result.size(); ++i) {
        cout << result[i] << endl;
    }
}

In Scala:

def printCoinCombinations(sum: Int, c1: Int, c2: Int, c3: Int) = {
    val sums = mutable.HashSet.empty[Int]
    sums.add(0)

    for (i <- 1 to sum) {
      if (sums.contains(i - c1) || sums.contains(i - c2) || sums.contains(i - c3)) {
        println(i)
        sums.add(i)
      }
    }
  }

Here is my C++ backtracking solution.

In summary in given vector of coins, we pick one element and deduct the coin value from the given target until we reach 0 or negative number.

class Solution {
 private:
    int sum;
    set<int> h;
 public:  
    void combinationSum(vector<int>& coins, int target) {     
      sum=0;
      dfs (candidates, 0, target);                
      for (auto t :h)
          cout<<t<<" ";
      return res;
    }
      void dfs (vector<int>& cand, int beg, int t) {
          if(t<=0) {
              return;
          }
          for(int i =beg ; i<cand.size();++i) {
              if (cand[i]<=t) {                  
                  sum+= cand[i];                      
                  h.insert(sum);
                  dfs (cand, i, t-cand[i]);
                  sum-=cand[i];
              }                  
          }
      }      
};

time complexity : n * 'search and insert of set'
= n * (2 * log n)
-> n log n
or
time complexity : log(1) + log(2) + ... + log(n)
= log(n!)
n/2 log(n/2) <= log(n!) <= n log n
so
O(n log n)

void printCoinsSum(const vector<int>& coins, int max_sum) {
    int start=coins[0];
    for(int i=1;i<coins.size();i++) {
        if(start>coins[i])
            start=coins[i];
    }

    set<int> coins_sum;
    coins_sum.insert(0);
    for(int sum=start;sum<=max_sum;sum++) {
        for(int i=0;i<coins.size();i++) {
            /*  new_sum = coins[i] + previous values
                new_sum - coins[i] = previous values
                check 'new_sum - coins[i]' is in previous values
            */
            if(coins_sum.find(sum-coins[i])!=coins_sum.end()) {
                coins_sum.insert(sum);
                break;
            }
        }
    }

    set<int>::const_iterator it=coins_sum.begin();
    ++it;
    for(;it!=coins_sum.end();it++)
        cout<<*it<<endl;
}

With PHP.
There is 3 possibilities to combine three different coins. Number can be valid between 1 to 1000 when get 0 that divided by any of three combination.

function possibleCombinedSums($c1, $c2, $c3){
		$combined1 = $c1 + $c2;
		$combined2 = $c1 + $c3;
		$combined3 = $c2 + $c3;

		for($sum = 1;$sum <= 1000;$sum++){
			if($sum%$combined1 == 0 || $sum%$combined2 == 0 || $sum%$combined3 == 0){
				print $sum.'<br/>';
			}

		}
	}

	possibleCombinedSums(5,10,15);

printSums(int[] coins) {
	Arrays.sort(coins)
	int minCoin = coins[0];

	boolean[] sums = new boolean[1000 - coins[0] + 1];

	for (int i = 0; i < coins.length; i++) {
		fillSums(sums, coins[i], minCoin);
	}

	for (int i = 0; i < sums.length; i++) {
		if (sums[i]) {
			
			system.out.println(i + minCoin);
		}
	}
}

fillSums(boolean[] arr, int coin, int minCoin) {
	arr[coin - minCoin] = true;

	for (int i = coin + 1; i - minCoin < arr.length; i++ ) {
		if ( (i- coin - minCoin) >= 0 && arr[i - coin - minCoin] ) {
			arr[i - minCoin] = true 
		}

	}
}

sums = range(1,1001)
numbers = [10, 15, 55]

new_sums = list(filter((lambda x: x%10 == 0 or x%15 == 0 or x%55 == 0), sums))
print(new_sums)

public coins(int a, int b, int c) {
  int step = gcd(a, b);
  step = gcd(step, c);
  int start = min(min(a, b), c);
  for (int i = start; i < 1000; i += step) {
    if (canFill(i, 0, a, b, c)) {
      System.out.println(i);
    }
  }
}

private boolean canFill(int needed, int my, int a, int b, int c) {
  if (my == needed) {
    return true;
  } else if (my > needed) {
    return false;
  } else {
    return canFill(needed, my + a, a, b, c) 
      || canFill(needed, my + b, a, b, c) 
      || canFill(needed, my + c, a, b, c);
  }
}

private int gcd(int a, int b) {
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}

A PreSum problem using HashSet or HashMap.
Why: nums adding up, similar to dp to store the previous value.

public int change(int amount, int[] coins) {
int[] combinations = new int[amount+1];
combinations[0]=1;

for(int coin : coins){
for(int i=1; i< combinations.length ;i++){
if(i>=coin){
combinations[i]+=combinations[i-coin];
}
}
}

return combinations[amount];
}

import java.util.ArrayList;
import java.util.Comparator;
import java.util.List;
import java.util.PriorityQueue;

public class Solution {
    private PriorityQueue<Integer> priorityQueue = new PriorityQueue<>();

    public ArrayList<Integer> intersect(List<Integer> denoms)
    {
        int sum = 0;

        denoms.sort(Comparator.comparingInt(o -> o));
        priorityQueue = new PriorityQueue<>();

        ArrayList<Integer> result = new ArrayList<>();
        while (sum < 1000) {
            for (int i = 0; i < denoms.size(); ++i) {
                priorityQueue.add(sum + denoms.get(i));
            }

            sum += priorityQueue.peek();

            if (sum < 1000) {
                result.add(sum);
            }
        }


        return result;
    }
}

def coin_three(a, b, c):
l = [a, b, c]
l.sort()
lout = [0]

index = 1
while True:
next_min = 2000
for i in l:
j = index-1
while True:
if j < 0 or lout[j] + i <= lout[index-1]:
break
if lout[j] + i < next_min:
next_min = lout[j] + i
j = j-1
lout.append(next_min)
print(next_min)
index += 1

if next_min >= 1000:
break

def coin_three(a, b, c):
    l = [a, b, c]
    l.sort()
    lout = [0]

    index = 1
    while True:
        next_min = 2000
        for i in l:
            j = index-1
            while True:
                if j < 0 or lout[j] + i <= lout[index-1]:
                    break
                if lout[j] + i < next_min:
                    next_min = lout[j] + i
                j = j-1
        lout.append(next_min)
        print(next_min)
        index += 1

        if next_min >= 1000:
            break

Keeping it simple with no datastructures or other nonsense

#include <stdio.h>
void coins_helper(a, b, c, na, nb, nc, total) {
	if ((a*na + b*nb+ c*nc) == total) {
		printf("%dx%d %dx%d %dx%d = %d\n", a, na, b, nb, c, nc, total);
		return;
	} else if ((a*na + b*nb+ c*nc) > total) {
    return;
  }
	
  coins_helper(a, b, c, na + 1, nb, nc, total);
	coins_helper(a, b, c, na, nb + 1, nc, total);
	coins_helper(a, b, c, na, nb, nc + 1, total);	
}

void coins(a, b, c, total) {
  coins_helper(a, b, c, 0, 0, 0, total);
}
int main(void) {
  coins(10, 25, 50, 100);
  return 0;
}

 ./main
10x10 25x0 50x0 = 100
10x5 25x2 50x0 = 100
10x5 25x0 50x1 = 100
10x5 25x2 50x0 = 100
10x5 25x2 50x0 = 100
10x5 25x0 50x1 = 100
10x5 25x2 50x0 = 100
10x5 25x2 50x0 = 100
10x5 25x2 50x0 = 100
10x5 25x0 50x1 = 100
10x5 25x2 50x0 = 100
10x5 25x2 50x0 = 100
10x5 25x2 50x0 = 100
10x5 25x2 50x0 = 100
10x5 25x0 50x1 = 100
10x5 25x2 50x0 = 100
10x5 25x2 50x0 = 100
10x5 25x2 50x0 = 100
10x5 25x2 50x0 = 100
10x5 25x2 50x0 = 100
10x5 25x0 50x1 = 100
10x5 25x2 50x0 = 100
10x5 25x2 50x0 = 100
10x5 25x2 50x0 = 100
10x5 25x2 50x0 = 100
10x5 25x2 50x0 = 100
10x5 25x2 50x0 = 100
10x0 25x4 50x0 = 100
10x0 25x2 50x1 = 100
10x0 25x2 50x1 = 100
10x5 25x0 50x1 = 100
10x0 25x2 50x1 = 100
10x0 25x0 50x2 = 100