How to find the minimum number of operation(s) to make the string balanced?

Question

From Codechef:

A string is considered balanced if and only if all the characters occur in it equal number of times.

You are given a strin

Answer 1

Following code implement the solution, in Java, with unit test.

The algorithm is almost identical as in @ruakh's answer, if not identical.

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Code

BalanceString.java

import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;

/**
 * Assume string only contains A-Z, the 26 uppercase letter,
 * 
given a string, you can replace a char with another char from the 26 letter,
 * 
figure out the minimum replacement required to make the string balance,
 * 
which means each char in the string occurs the same time,
 *
 * @author eric
 * @date 2/2/19 8:54 PM
 */
public class BalanceString {
    private final char minChar;
    private final char maxChar;
    private final int distinctChars; // total distinct char count,

    public static final BalanceString EN_UPPER_INSTANCE = new BalanceString('A', 'Z');

    public BalanceString(char minChar, char maxChar) {
        this.minChar = minChar;
        this.maxChar = maxChar;
        this.distinctChars = maxChar - minChar + 1;

        if (distinctChars <= 0)
            throw new IllegalArgumentException("invalid range of chars: [" + minChar + ", " + maxChar + "]");
    }

    /**
     * Check minimal moves needed to make string balanced.
     *
     * @param str
     * @return count of moves
     */
    public int balanceCount(String str) {
        // System.out.printf("string to balance:\t%s\n", str);
        int len = str.length(); // get length,
        if (len <= 2) return 0; // corner cases,

        Map coMap = figureOccurs(str); // figure occurrences,
        Integer[] occurs = sortOccursReversely(coMap); // reversely order occurrences,

        int m = coMap.size(); // distinct char count,
        int maxN = (len < distinctChars ? len : distinctChars); // get max possible distinct char count, included,

        int smallestMoves = Integer.MAX_VALUE; // smallest moves, among all possible n,

        // check each possible n, and get its moves,
        for (int n = 1; n <= maxN; n++) {
            if (len % n == 0) {
                int moves = figureMoves(len, coMap, occurs, m, n);
                if (moves < smallestMoves) smallestMoves = moves;
            }
        }

        return smallestMoves;
    }

    /**
     * Figure occurs for each char.
     *
     * @param str
     * @return
     */
    protected Map figureOccurs(String str) {
        Map coMap = new HashMap<>();
        for (char c : str.toCharArray()) {
            if (c < minChar || c > maxChar)
                throw new IllegalArgumentException(c + " is not within range 'A-Z'");

            if (!coMap.containsKey(c)) coMap.put(c, 1);
            else coMap.put(c, coMap.get(c) + 1);
        }

        return coMap;
    }

    /**
     * Get reverse sorted occurrences.
     *
     * @param coMap
     * @return
     */
    protected Integer[] sortOccursReversely(Map coMap) {
        Integer[] occurs = new Integer[coMap.size()];

        coMap.values().toArray(occurs);
        Arrays.sort(occurs, Collections.reverseOrder());

        return occurs;
    }

    /**
     * Figure moves needed to balance.
     *
     * @param len   length of string,
     * @param coMap
     * @param m     original distinct elements count,
     * @param n     new distinct elements count,
     * @return count of moves,
     */
    protected int figureMoves(int len, Map coMap, Integer[] occurs, int m, int n) {
        int avgOccur = len / n; // average occurrence,
        int moves = 0;

        if (n == m) { // distinct chars don't change,
            for (Integer occur : occurs) {
                if (occur <= avgOccur) break;
                moves += (occur - avgOccur);
            }
        } else if (n < m) { // distinct chars decrease,
            for (int i = 0; i < n; i++) moves += Math.abs(occurs[i] - avgOccur); // for elements kept,
            for (int i = n; i < m; i++) moves += occurs[i]; // for elements to replace,
            moves >>= 1;
        } else { // distinct chars increase,
            for (int i = 0; i < occurs.length; i++) moves += Math.abs(occurs[i] - avgOccur); // for existing elements,
            moves += ((n - m) * avgOccur); // for new elements,
            moves >>= 1;
        }

        return moves;
    }

    public char getMinChar() {
        return minChar;
    }

    public char getMaxChar() {
        return maxChar;
    }

    public int getDistinctChars() {
        return distinctChars;
    }
}

BalanceStringTest.java
(Unit test, via TestNG)

import org.testng.Assert;
import org.testng.annotations.Test;

/**
 * BalanceString test.
 *
 * @author eric
 * @date 2/2/19 9:36 PM
 */
public class BalanceStringTest {
    private BalanceString bs = BalanceString.EN_UPPER_INSTANCE;

    @Test
    public void test() {
        // n < m case,
        Assert.assertEquals(bs.balanceCount("AAAABBBC"), 1); // e.g 1A -> B,
        Assert.assertEquals(bs.balanceCount("AAAAABBC"), 2); // e.g 1A -> B, 1C -> B,
        Assert.assertEquals(bs.balanceCount("AAAAAABC"), 2); // e.g 1B -> A, 1C -> A,
        Assert.assertEquals(bs.balanceCount("AAAAAAAB"), 1); // e.g 1B -> A,

        // n > m case,
        Assert.assertEquals(bs.balanceCount("AAAABBBBCCCCDDDDEEEEAAAA"), 4); // add 1 new char, e.g change 4 A to 4 F,
        Assert.assertEquals(bs.balanceCount(genIncSeq(10)), 15); // A-J, 10 distinct chars, 55 in length; solved by add 1 new char, need 15 steps,

        // n == m case,
        Assert.assertEquals(bs.balanceCount(genIncSeq(3)), 1); // A-C, 3 distinct chars, 6 in length; symmetric, solved with same distinct chars, need 1 steps,
        Assert.assertEquals(bs.balanceCount(genIncSeq(11)), 15); // A-K, 11 distinct chars, 66 in length; symmetric, solved with same distinct chars, need 15 steps,

        // n < m, or n > m case,
        Assert.assertEquals(bs.balanceCount("ABAC"), 1); // e.g 1A -> B, or 1A -> D,
    }

    // corner case,
    @Test
    public void testCorner() {
        // m <= 2,
        Assert.assertEquals(bs.balanceCount(""), 0);
        Assert.assertEquals(bs.balanceCount("A"), 0);
        Assert.assertEquals(bs.balanceCount("AB"), 0);
        Assert.assertEquals(bs.balanceCount("AA"), 0);

        /*------ m == n == distinctChars ------*/
        String mndBalanced = genMndBalancedSeq(); // each possible char occurs exactly once, already balanced,
        Assert.assertEquals(mndBalanced.length(), bs.getDistinctChars());
        Assert.assertEquals(bs.balanceCount(mndBalanced), 0); // no need change,

        char lastChar = mndBalanced.charAt(mndBalanced.length() - 1);
        String mndOneDup = mndBalanced.replace(lastChar, (char) (lastChar - 1)); // (distinctChars -2) chars occur exactly once, one occurs twice, one is missing, thus it's one step away to balance,
        Assert.assertEquals(mndOneDup.length(), bs.getDistinctChars());
        Assert.assertEquals(bs.balanceCount(mndOneDup), 1); // just replace the duplicate char with missing char,
    }

    // invalid input,
    @Test(expectedExceptions = IllegalArgumentException.class)
    public void testInvalidInput() {
        Assert.assertEquals(bs.balanceCount("ABAc"), 1);
    }

    // invalid char range, for constructor,
    @Test(expectedExceptions = IllegalArgumentException.class)
    public void testInvalidRange() {
        new BalanceString('z', 'a');
    }

    /**
     * Generate a string, with first char occur once, second twice, third three times, and so on.
     * 
e.g A, ABB, ABBCCC, ABBCCCDDDD,
     *
     * @param m distinct char count,
     * @return
     */
    private String genIncSeq(int m) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < m; i++) {
            for (int j = 0; j <= i; j++) sb.append((char) (bs.getMinChar() + i));
        }
        return sb.toString();
    }

    /**
     * Generate a string that contains each possible char exactly once.
     * 
For [A-Z], it could be: "ABCDEFGHIJKLMNOPQRSTUVWXYZ".
     *
     * @return
     */
    private String genMndBalancedSeq() {
        StringBuilder sb = new StringBuilder();
        char minChar = bs.getMinChar();
        int distinctChars = bs.getDistinctChars();

        for (int i = 0; i < distinctChars; i++) {
            sb.append((char) (minChar + i));
        }
        return sb.toString();
    }
}

All test cases would pass.

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Complexity

• Time: O(len) + O(m * lg(m)) + O(m * factorCount)
  • Each sequential scan takes O(len), there are several sequential loop.
  • Sorting of array takes O(m*lg(m)), which is at most O(distinctChars * lg(distinctChars)), thus constant, and only sort once.
  • To figure out moves for each n, takes O(m).
  • The count of n that needs to figure moves, depends on count of divisible numbers for len, in the range [minChar, maxChar].
    This count if kind small & constant too.
• Space: O(len)
  • Input string need O(len).
  • Counter hashmap need O(m).
  • Sorted occurrence array need O(m).

Where:

• len is string length.
• m is distinct char count in original string
• distinctChars is distinct char count, e.g 26.
• maxN max possible distinct char count, included,
• factorCount divisible number count in range [1, n], by len,
• minChar min char, e.g A
• maxChar max char, e.g Z

And:

• len >= m
• m <= distinctChars