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Home/ Questions/Q 8812023
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T03:30:31+00:00 2026-06-14T03:30:31+00:00

Why does containsAll method on a HashSet does not remain consistent if remove is

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Why does containsAll method on a HashSet does not remain consistent if remove is called on the Set whereas a containsValue method on a HashMap remains consistent after a value is removed
After a value is removed from a HashSet containsAll returns false even if all values were present where as in case of HashMap the containsValue method returns correct value

public static void main(String[] args)
    {

    HashSet<String> lookup=new HashSet<String>();
    HashMap<Integer,String> findup=new HashMap<Integer,String>();

    String[] Alltokens={"This","is","a","programming","test","This","is","a","any","language"};
    for(String s:Alltokens)
        lookup.add(s);

    String[] tokens={"This","is","a"};
    Collection<String> tok=Arrays.asList(tokens);

    lookup.remove(Alltokens[5]); 
     if(lookup.containsAll(tok))
       System.out.print("found");
    else    
        System.out.print("NOT found");

    Integer i=0;
    for(String s:Alltokens)
        findup.put(i++,s);
    boolean found=true;
    findup.remove(Alltokens[0]);
        findup.remove(5);               //Edit : trying to remove value whose key is 5
    for(String s:tokens)
        if(!findup.containsValue(s))
            found=false;
    System.out.print("\nIn map " + found);
}

Output
NOT found
In map true

Is there a way to keep containsAll consistent if remove method is called on the HashSet?
Also if a value that was not present in the set is passed to remove method.ContainsAll remains consistent

        lookup.remove("Alltokens[5]");
    if(lookup.containsAll(tok))

//This will be true now where as it is false if a value already present is removed

May be it has got to do something with keys in HashMaps and no keys in HashSet.Can you please explain how do they work?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-14T03:30:32+00:00Added an answer on June 14, 2026 at 3:30 am

    Map.remove(Object) removes a mapping based on the key.

    Since you use Integer objects as the keys (put(i++, s)) you will remove nothing when you call findup.remove(Alltokens[0])! Check its return value to see that it will return false.

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