Coin change III - count distinct total ways to make combination

shriyanshi24jindal

Mar 28th, 2026

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Java 20.70 KB | None | 0 0

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class Solution {
// idea: Do not add +1 when picking a coin. Each valid combination is counted exactly once.
/** Recursion */
public int change(int amount, int[] coins) {
int n = coins.length;
return findWays(coins, amount, n);
}
public int findWays(int[] coins, int amount, int n)
{
// base condition: when amount becomes 0 then we found valid combination
if(amount == 0)
return 1;
// base condition: No coins left
// At this point, if amount > 0 (positive), we cannot form the remaining amount, because there are no coins left to use.
// Returning 0 here would mean 0 coins are needed, which is incorrect, because we cannot form the positive amount with no coins.
if(n == 0)
return 0;
if(coins[n-1] <= amount)
{
int pick = findWays(coins, amount-coins[n-1], n);
int notPick = findWays(coins, amount, n-1);
return pick + notPick;
}
return findWays(coins, amount, n-1);
}
/** Memoization */
public int change(int amount, int[] coins) {
int n = coins.length;
// create a matrix
int[][] t = new int[n+1][amount+1];
// initialise the matrix
for(int i=0; i<n+1; i++)
Arrays.fill(t[i], 1);
for(int j=0; j<n+1; j++)
Arrays.fill(t[j], 0);
return findWays(coins, amount, n, t);
}
public int findWays(int[] coins, int amount, int n, int[][] t)
{
// base condition: when amount becomes 0 then we found valid combination
if(amount == 0)
return 1;
// base condition: No coins left
// At this point, if amount > 0 (positive), we cannot form the remaining amount, because there are no coins left to use.
// Returning 0 here would mean 0 coins are needed, which is incorrect, because we cannot form the positive amount with no coins.
if(n == 0)
return 0;
if(t[n][amount] != 0)
return t[n][amount];
if(coins[n-1] <= amount)
{
int pick = findWays(coins, amount-coins[n-1], n, t);
int notPick = findWays(coins, amount, n-1, t);
t[n][amount] = pick + notPick;
return t[n][amount];
}
t[n][amount] = findWays(coins, amount, n-1, t);
return t[n][amount];
}
/** Top-down */
public int change(int amount, int[] coins) {
int n = coins.length;
// create a matrix
int[][] t = new int[n+1][amount+1];
// initialise the matrix with base condition
for(int i=0; i<n+1; i++)
{
for(int j=0; j<amount+1; j++)
{
// when we reach end of the array
if(i == 0)
t[i][j] = 0;
// when sum is always 0, then empty subset
if(j == 0)
t[i][j] = 1;
}
}
for(int i=1; i<n+1; i++)
{
for(int j=1; j<amount+1; j++)
{
if(coins[i-1] <= j)
{
int pick = t[i][j - coins[i-1]];
int notPick = t[i-1][j];
t[i][j] = pick + notPick;
}
else
t[i][j] = t[i-1][j];
}
}
return t[n][amount];
}
}

RAW Paste Data Copied

class Solution {
    // idea: Do not add +1 when picking a coin. Each valid combination is counted exactly once.

/** Recursion */
    public int change(int amount, int[] coins) {
        int n = coins.length;

return findWays(coins, amount, n);
    }

public int findWays(int[] coins, int amount, int n)
    {
        // base condition: when amount becomes 0 then we found valid combination
        if(amount == 0)
            return 1;

// base condition: No coins left
        // At this point, if amount > 0 (positive), we cannot form the remaining amount, because there are no coins left to use.
        // Returning 0 here would mean 0 coins are needed, which is incorrect, because we cannot form the positive amount with no coins.
        if(n == 0)
            return 0;
        
        if(coins[n-1] <= amount)
        {
            int pick = findWays(coins, amount-coins[n-1], n);
            int notPick = findWays(coins, amount, n-1);
            return pick + notPick;
        }
        return findWays(coins, amount, n-1);
    }

/** Memoization */
    public int change(int amount, int[] coins) {
        int n = coins.length;

// create a matrix
        int[][] t = new int[n+1][amount+1];

// initialise the matrix
        for(int i=0; i<n+1; i++)
            Arrays.fill(t[i], 1);

for(int j=0; j<n+1; j++)
            Arrays.fill(t[j], 0);

return findWays(coins, amount, n, t);
    }

public int findWays(int[] coins, int amount, int n, int[][] t)
    {
        // base condition: when amount becomes 0 then we found valid combination
        if(amount == 0)
            return 1;

// base condition: No coins left
        // At this point, if amount > 0 (positive), we cannot form the remaining amount, because there are no coins left to use.
        // Returning 0 here would mean 0 coins are needed, which is incorrect, because we cannot form the positive amount with no coins.
        if(n == 0)
            return 0;
    
        if(t[n][amount] != 0)
            return t[n][amount];
        
        if(coins[n-1] <= amount)
        {
            int pick = findWays(coins, amount-coins[n-1], n, t);
            int notPick = findWays(coins, amount, n-1, t);
            t[n][amount] = pick + notPick;
            return t[n][amount];
        }
        t[n][amount] = findWays(coins, amount, n-1, t);
        return t[n][amount];
    }

/** Top-down */
    public int change(int amount, int[] coins) {
        int n = coins.length;

// create a matrix
        int[][] t = new int[n+1][amount+1];

// initialise the matrix with base condition
        for(int i=0; i<n+1; i++)
        {
            for(int j=0; j<amount+1; j++)
            {
                // when we reach end of the array
                if(i == 0)
                    t[i][j] = 0;
                
                // when sum is always 0, then empty subset
                if(j == 0)
                    t[i][j] = 1;
            }
        }

for(int i=1; i<n+1; i++)
        {
            for(int j=1; j<amount+1; j++)
            {
                if(coins[i-1] <= j)
                {
                    int pick = t[i][j - coins[i-1]];
                    int notPick = t[i-1][j];
                    t[i][j] = pick + notPick;
                }
                else
                    t[i][j] = t[i-1][j];
            }
        }
        return t[n][amount];
    }
}