# Stickers to Spell Word

This page explains Java solution to problem `Stickers to Spell Word` using `Dynamic Programming` algorithm.

## Problem Statement

We are given `N` different types of stickers. Each sticker has a lowercase English word on it.

You would like to spell out the given target string by cutting individual letters from your collection of stickers and rearranging them.

You can use each sticker more than once if you want, and you have infinite quantities of each sticker.

What is the minimum number of stickers that you need to spell out the target? If the task is impossible, return `-1`.

Example 1:

Input: ["with", "example", "science"], "thehat"
Output: 3

Example 2:

Input: ["notice", "possible"], "basicbasic"
Output: -1

## Solution

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``````
package com.vc.hard;

class StickersToSpellWord {
public int minStickers(String[] stickers, String target) {
/**
target: hat
stickers: haha
tata

possible states dp[1 << hat.length()]
hat -> final state 111 and 000 as initial state

iteration 1:
000 -> apply first sticker 110, apply second sticker 011
001 -> -1
010 -> -1
011 -> one sticker of tata
100 -> -1
101 -> -1
110 -> one sticker haha
111 -> -1

iteration 2: find next non "-1" state which is 011 and apply stickers
000 -> apply first sticker 110, apply second sticker 011
001 -> -1
010 -> -1
011 ->  1(tata) applying haha
100 -> -1
101 -> -1
110 ->  1(haha)
111 ->  2(haha + tata)

And so on ...
*/
int n = target.length();
int[] dp = new int[1 << n];
for(int i = 1; i < (1 << n); i++) dp[i] = -1;

for(int state = 0; state < (1 << n); state++) {
if(dp[state] == -1) continue;
for(String sticker: stickers) {
int initialState = state;
for(char ch: sticker.toCharArray()) {
for(int i = 0; i < n; i++) {
/** This bit is already set */
if (((initialState >> i) & 1) == 1) continue;
if(target.charAt(i) == ch) {
initialState |= (1 << i);
break; /** Break, as we have used current character from the sticker */
}
}
}
if(dp[initialState] == -1 || dp[initialState] > dp[state] + 1) {
dp[initialState] = dp[state] + 1;
}
}
}
return dp[(1 << n) - 1];
}
}
``````

## Time Complexity

O(2T * N * T) Where
N is total type of stickers in an input string array
T is length of target string

## Space Complexity

O(2T) Where
T is length of target string